Inter 1st Year

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 8 Limits and Continuity to solve questions creatively.

Intermediate 1st Year Maths 1B Limits and Continuity Formulas

Right Limit :
Suppose f is defined on (a, b) and Z ∈ R. Given ε < 0 , there exists δ > 0 such that a < x < a + δ => |f(x) – l| < ε, then l is said to be the right limit of’ f’ at ‘a’.
It is denoted by \({Lt}_{x \rightarrow a+} f(x)\)f(x) = l

Left Limit :
Suppose ‘ f’ is defined on (a, b) and Z e R. Given e > 0, there exists δ > 0 such that a – δ < x < a ⇒ |f(x) – l| < ε, then l is said to be the left limit of’ f’ at ’a’ and is denoted by \({Lt}_{x \rightarrow a-} \dot{f(x)}\) = l

Suppose f is defined in a deleted neighbourhood of a and l ∈ R
\({Lt}_{x \rightarrow a} f(x)=l \Leftrightarrow \underset{x \rightarrow a+}{L t} f(x)={Lt}_{x \rightarrow a-} \quad f(x)=l\)

Standard limits:

• \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}\) = na^n-1
• \({Li}_{x \rightarrow 0} \frac{\sin x}{x}\) = 1 (x is in radians)
• \({Lt}_{x \rightarrow 0} \frac{\tan x}{x}\) = 1
• \({lt}_{x \rightarrow 0}(1+x)^{1 / x}\) = e
• \({Lit}_{x \rightarrow 0}\left(1+\frac{1}{x}\right)^{x}\) = e
• \({Lt}_{x \rightarrow 0}\left(\frac{a^{x}-1}{x}\right)\) = log_ea
• \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x^{m}-a^{m}}=\frac{n}{m}\)a^{n – m}
• \({Lt}_{x \rightarrow 0} \frac{\sin a x}{x}\) = a(x is in radians)
• \({Lt}_{x \rightarrow 0} \frac{\tan a x}{x}\) = a(x is in radians)
• \({Lt}_{x \rightarrow \infty}\left(1+\frac{p}{x}\right)^{Q x}\) = e^PQ
• \({lt}_{x \rightarrow 0} \frac{e^{x}-1}{x}\) = 1
• \({Lt}_{x \rightarrow 0}(1+p x)^{\frac{Q}{x}}\) = e^PQ

Intervals
Definition:
Let a, b ∈ R and a < b. Then the set {x ∈ R: a ≤ x ≤ b} is called a closed interval. It is denoted by [a, b]. Thus
Closed interval [a, b] = {x ∈ R: a ≤ x ≤ b}. It is geometrically represented by

Open interval (a,b) = {x ∈ R: a < x < b} It is geometrically represented by

Left open interval
(a, b] = {x ∈ R: a < x ≤ b}. It is geometrically represented by

Right open interval
[a, b) = {x ∈ R: a ≤ x < b}. It is geometrically represented by

[a, ∞) = {x ∈ R : x ≥ a} = {x ∈ R : a ≤ x < ∞} It is geometrically represented by

(a, ∞) = {x ∈ R : x > a} = {x ∈ R : a < x < ∞}

(-∞, a] = {x ∈ R : x ≤ a} = {xe R : -∞ < x < a}

Neighbourhood of A Point: Definition: Let ae R. If δ > 0 then the open interval (a – δ, a + δ) is called the neighbourhood (δ – nbd) of the point a. It is denoted by N_δ (a) . a is called the centre and δ is called the radius of the neighbourhood .
∴ N_δ(a) = (a – δ, a + δ) = {x ∈ R: a – δ< x < a + δ} = {x ∈ R: |x – a| < δ}

The set N_δ(a) – {a} is called a deleted δ – neighbourhood of the point a.
∴ N_δ(a) – {a} = (a – δ, a) ∪ (a, a + δ) = {x ∈ R :0 < | x – a | < δ}
Note: (a – δ, a) is called left δ -neighbourhood, (a, a + δ) is called right δ – neighbourhood of a

Graph of A Function:

Mod function:
The function f: R-R defined by f(x) = |x| is called the mod function or modulus function or absolute value function.
Dom f R. Range f [0, )

Reciprocal function :
The function f: R – {0} – R defined by
f(x) = \(\frac{1}{x}\) is called the reciprocal function,
Dom f = R – {0}= Range f = R

Identity function:
The function f R-R defined by f(x) = x is called the identity
It is denoted by I(x)

Limit of A Function
Concept of limit:
Before giving the formal definition of limit consider the following example.
Let f be a function defined by f (x) = \(\frac{x^{2}-4}{x-2}\). clearly, f is not defined at x = 2.
When x ≠ 2, x – 2 ≠ 0 and f(x) = \(\frac{(x-2)(x+2)}{x-2}\) = x + 2

Now consider the values of f(x) when x ≠ 2, but very very close to 2 and <2.

It is clear from the above table that as x approaches 2 i.e.,x → 2 through the values less than 2, the value of f(x) approaches 4 i.e., f(x) → 4. We will express this fact by saying that left hand limit of f(x) as x → -2 exists and is equal to 4 and in symbols we shall write \({lt}_{x \rightarrow 2^{-}} f(x)\) = 4

Again we consider the values of f(x) when x ≠ 2, but is very-very close to 2 and x > 2.

It is clear from the above table that as x approaches 2 i.e.,x → 2 through the values greater than 2, the value of f(x) approaches 4 i.e., f(x) → 4. We will express this fact by saying that right hand
limit of f(x) as x → 2 exists and is equal to 4 and in symbols we shall write \({lt}_{x \rightarrow 2^{+}} f(x)\) = 4

Thus we see that f(x) is not defined at x = 2 but its left hand and right hand limits as x → 2 exist and are equal.
When \({lt}_{\mathrm{x} \rightarrow \mathrm{a}^{+}} \mathrm{f}(\mathrm{x}), \mathrm{lt}_{\mathrm{x} \rightarrow \mathrm{a}^{-}}^{\mathrm{f}(\mathrm{x})}\) are equal to the same number l, we say that \(\begin{array}{ll}
l_{x \rightarrow a} & f(x)
\end{array}\) exist and equal to 1.

Thus, in above example,

One Sided Limits Definition Of Left Hand Limit:
Let f be a function defined on (a – h, a), h > 0. A number l1 is said to be the left hand limit (LHL) or left limit (LL) of f at a if to each
ε >0, ∃ a δ >0 such that, a – δ < x < a ⇒ |f (x) – l₁| < ε.
In this case we write \(\underset{x \rightarrow a-}{L t} f(x)\) = l₁ (or) \({Lt}_{x \rightarrow a-0} f(x)\) = l₁

Definition of Right Limit:
Let f be a function defined on (a, a + h), h > 0. A number l ₂is said to the right hand limit (RHL) or right limit (RL) of f at a if to each ε >0, ∃ a δ >0 such that, a – δ < x < a ⇒ |f (x) – l₂| < ε.
In this case we write \(\underset{x \rightarrow a+}{L t} f(x)\) = l₂ (or) \({Lt}_{x \rightarrow a+0} f(x)\) = l₂

Definition of Limit:
Let A ∈ R, a be a limit point of A and
f : A → R. A real number l is said to be the limit of f at a if to each ε > 0, ∃ a δ > 0 such that x ∈ A, 0 < |x – a| < δ ⇒ | f(x) – l| < ε. In this case we write f (x) → 2 (or) \({Lt}_{x \rightarrow a} f(x)\) = l. Note: 1.If a function f is defined on (a – h, a) for some h > 0 and is not defined on (a, a + h) and if \(\underset{x \rightarrow a-}{{Lt}} f(x)\) exists then \({Lt}_{x \rightarrow a} f(x)={Lt}_{x \rightarrow a-} f(x)\).
2. If a function f is defined on (a, a + h) for some h > 0 and is not defined on (a – h, a) and if \(\underset{x \rightarrow a+}{{Lt}} f(x)\) exists then \({Lt}_{x \rightarrow a} f(x)={Lt}_{x \rightarrow a+} f(x)\).

Theorem:
If \({Lt}_{x \rightarrow a} f(x)\) exists then \({Lt}_{x \rightarrow a} f(x)={Lt}_{x \rightarrow 0} f(x+a)={Lt}_{x \rightarrow 0} f(a-x)\)

Theorems on Limits WithOut Proofs
1. If f : R → R defined by f(x) = c, a constant then \({Lt}_{x \rightarrow a} f(x)\) = c for any a ∈ R.

2. If f: R → R defined by f(x) = x, then \({Lt}_{x \rightarrow a} f(x)\) = a i.e., \({Lt}_{x \rightarrow a} f(x)\) = a (a ∈ R)

3. Algebra of limits

vii) If f(x) ≤ g(x) in some deleted neighbourhood of a, then \({Lt}_{x \rightarrow a} f(x) \leq {Lt}_{x \rightarrow a} g(x)\)
viii) If f(x) ≤ h(x) < g(x) in a deleted nbd of a and \({Lt}_{x \rightarrow a} f(x)\) = l = \({Lt}_{x \rightarrow a} g(x)\) then \({Lt}_{x \rightarrow a} h(x)\) = l
ix) If \({Lt}_{x \rightarrow a} f(x)\) = 0 and g(x) is a bounded function in a deleted nbd of a then \({Lt}_{x \rightarrow a}\) f(x) g(x) = 0.

Theorem
If n is a positive integer then \({Lt}_{x \rightarrow a} x^{n}\) = aⁿ, a ∈ R

Theorem
If f(x) is a polynomial function, then \({Lt}_{x \rightarrow a}\)f(x) = f (a)

Evaluation of Limits:
Evaluation of limits involving algebraic functions.
To evaluate the limits involving algebraic functions we use the following methods:

• Direct substitution method
• Factorisation method
• Rationalisation method
• Application of the standard limits.

1. Direct substitution method:
This method can be used in the following cases:

• If f(x) is a polynomial function, then \({Lt}_{x \rightarrow a}\)f(x) = f (a).
• If f (x) = \(\frac{P(a)}{Q(a)}\) where P(x) and Q(x) are polynomial functions then \({Lt}_{x \rightarrow a}\)f(x) = \(\frac{P(a)}{Q(a)}\) provided Q(a) ≠ 0.

2. Factiorisation Method:
This method is used when \({Lt}_{x \rightarrow a}\)f (x) is taking the indeterminate form of the type 0 by the substitution of x = a.
In such a case the numerator (Nr.) and the denominator (Dr.) are factorized and the common factor (x – a) is cancelled. After eliminating the common factor the substitution x = a gives the limit, if it exists.

3. Rationalisation Method : This method is used when \({Lt}_{x \rightarrow a} \frac{f(x)}{g(x)}\) is a \(\frac{0}{0}\) form and either the Nr. or Dr. consists of expressions involving radical signs.

4. Application of the standard limits.
In order to evaluate the given limits , we reduce the given limits into standard limits form and then we apply the standard limits.

Theorem 1.
If n is a rational number and a > 0 then \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}\) = n.a^n-1

Note:

• If n is a positive integer, then for any a ∈ R. \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}\) = n.a^n-1
• If n is a real number and a> 0 then \({Lt}_{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}\) = n.a^n-1
• If m and n are any real numbers and a > 0, then \({Lt}_{x \rightarrow a} \frac{x^{m}-a^{m}}{x^{n}-a^{n}}=\frac{m}{n}\)a^m-n

Theorem 2.
If 0 < x < \(\frac{\pi}{2}\) then sin x < x < tan x.

Corollary 1:
If – \(\frac{\pi}{2}\) < x < 0 then tan x < x < sin x

Corollary 2:
If 0 < |x| < \(\frac{\pi}{2}\) then |sin x| < |x| < |tan x|

Standard Limits:

• \({Lt}_{x \rightarrow 0} \frac{\sin x}{x}\) = 1
• \(\underset{x \rightarrow 0}{L t} \frac{\tan x}{x}\) = 1
• \({Lt}_{x \rightarrow 0} \frac{e^{x}-1}{x}\) = 1
• \({Lt}_{x \rightarrow 0} \frac{a^{x}-1}{x}\) = log_ea
• \({Lt}_{x \rightarrow 0}(1+x)^{\frac{1}{x}}\) = e
• \({Lt}_{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}\) = e
• \({Lt}_{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}\) = e

Limits At Infinity
Definition:
Let f(x) be a function defined on A = (K,).
(i) A real number l is said to be the limit of f(x) at ∞ if to each δ > 0, ∃ , an M > 0 (however large M may be) such that x ∈ A and x > M ⇒ |f (x) – l|< δ.
In this case we write f (x) → l as x → +∞ or \({Lt}_{x \rightarrow \infty} f(x)\) = l.

(ii) A real number l is said to be the limit of f(x) at -∞ if to each δ > 0, ∃ > 0, an M > 0 (however large it may be) such that x ∈ A and x < – M ⇒ |f (x) – l| < δ. In this case, we write f (x) → l as x → -∞ or \({Lt}_{x \rightarrow \infty} f(x)\) = l.

Infinite Limits Definition:
(i) Let f be a function defined is a deleted neighbourhood of D of a. (i) The limit of f at a is said to be ∞ if to each M > 0 (however large it may be) a δ > 0 such that x ∈ D, 0 < |x – a| < δ ⇒ f(x) > M. In this case we write f(x) as x → a or \({Lt}_{x \rightarrow a} f(x)\) = +∞

(ii) The limit of f(x) at a is said be -∞ if to each M > 0 (however large it may be) a δ > 0 such that x ∈ D, 0 < | x – a | < δ ⇒ f(x) < – M. In thise case we write f(x) → -∞ as x → a or \({Lt}_{x \rightarrow a} f(x)\) = -∞

Indeterminate Forms:
While evaluation limits of functions, we often get forms of the type \(\frac{0}{0}, \frac{\infty}{\infty}\), 0 × -∞, 0⁰, 1^∞, ∞⁰ which are termed as indeterminate forms.

Continuity At A Point:
Let f be a function defined in a neighbourhood of a point a. Then f is said to be continuous at the point a if and only if \({Lt}_{x \rightarrow a} f(x)\) = f (a).
In other words, f is continuous at a iff the limit of f at a is equal to the value of f at a.

Note:

• If f is not continuous at a it is said to be discontinuous at a, and a is called a point of discontinuity of f.
• Let f be a function defined in a nbd of a point a. Then f is said to be
  • Left continuous at a iff \({Lt}_{x \rightarrow a-} f(x)\) = f (a).
  • Right continuous at a iff \({Lt}_{x \rightarrow a+} f(x)\) = f (a).
• f is continuous at a iff f is both left continuous and right continuous at a
  i.e, \({Lt}_{x \rightarrow a} f(x)=f(a) \Leftrightarrow {Lt}_{x \rightarrow a^{-}} f(x)=f(a)={Lt}_{x \rightarrow a+} f(x)\)

Continuity of A Function Over An Interval:
A function f defined on (a, b) is said to be continuous (a,b) if it is continuous at everypoint of (a, b) i.e., if \({Lt}_{x \rightarrow c} f(x)\) =f (c) ∀c ∈ (a, b)
II) A function f defined on [a, b] is said to be continuous on [a, b] if

• f is continuous on (a, b) i.e., \({Lt}_{x \rightarrow c} f(x)\) = f (c) ∀c ∈(a, b)
• f is right continuous at a i.e., \({Lt}_{x \rightarrow a+} f(x)\) = f (a)
• f is left continuous at b i.e., \({Lt}_{x \rightarrow b-} f(x)\) = f (b).

Note :

• Let the functions f and g be continuous at a and k€R. Then f + g, f – g, kf , kf + lg, f.g are continuous at a and \(\frac{f}{g}\) is continuous at a provided g(a) ≠ 0.
• All trigonometric functions, Inverse trigonometric functions, hyperbolic functions and inverse hyperbolic functions are continuous in their domains of definition.
• A constant function is continuous on R
• The identity function is continuous on R.
• Every polynomial function is continuous on R.

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 4 Pair of Straight Lines to solve questions creatively.

Intermediate 1st Year Maths 1B Pair of Straight Lines Formulas

→ If ax² + 2hxy + by² = 0 represents a pair of lines, then the sum of the slopes of lines is \(-\frac{2 h}{b}\) and the product of the slopes of lines is \(\frac{a}{b}\).

→ If ‘θ’ is an angle between the lines represented by ax² + 2hxy + by² = 0
then cos θ = \(\frac{a+b}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
tan θ = \(\frac{2 \sqrt{h^{2}-a b}}{a+b}\)
If ‘θ’ is accute, cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}\); tan θ = \(\frac{2 \sqrt{h^{2}-a b}}{|a+b|}\)

→ If h² = ab, then ax² + 2hxy + by² = 0 represents coincident or parallel lines.

→ ax² + 2hxy + by² = 0 represents a pair of ⊥^lr lines ⇒ a+ b = 0 i.e., coeft. of x² + coeff. of y² = 0.

→ The equation of pair of lines passing through origin and perpendicular to ax² + 2hxy + by² = 0 is bx² – 2hxy + ay² = 0.

→ The equation of pair of lines passing through (x₁, y₁) and perpendicular to ax² + 2hxy + by² = 0 isb (x – x₁)² – 2h(x – x₁) (y – y₁) + a(y – y₁)² = 0.

→ The equation of pair of lines passing through (x₁, y₁) and parallel to ax² + 2hxy + by² = 0 is a(x-x,)² + 2h(x-x1)(y-y1) + b(y – y₁)² = 0.

→ The equations of bisectors of angles between the lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, is \(\frac{a_{1} x+b_{1} y+c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\pm \frac{\left(a_{2} x+b_{2} y+c_{2}\right)}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\)

→ The equation to the pair of bisectors of angles between the pair of lines ax² + 2hxy + by² = 0 is h(x² – y²) = (a – b)xy.

→ The product of the perpendicular is from (α, β) to the pair of lines ax² + 2hxy + by² = 0 is \(\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

→ The area of the triangle formed by ax² + 2hxy + by² = 0 and lx + my + n = 0 is \(\frac{n^{2} \sqrt{h^{2}-a b}}{\left|a m^{2}-2 h / m+b\right|^{2} \mid}\)

→ The line ax + by + c = 0 and pair of lines (ax + by)² – 3(bx – ay)² =0 form an equilateral triangle and the area is \(\frac{c^{2}}{\sqrt{3}\left(a^{2}+b^{2}\right)}\) sq.units

→ If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines, then

• abc + 2fgh – af² – bg² – ch² = 0
• h² ≥ ab
• g² ≥ ac
• f² ≥ be

→ If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines and h² > ab, then the point of intersection of the lines is \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

→ If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of parallel lines then h² = ab and af² = bg²
The distance between the parallel lines = \(2 \sqrt{\frac{g^{2}-a c}{a(a+b)}}=2 \sqrt{\frac{f^{2}-b c}{b(a+b)}}\)

Pair of Straight Lines:
Let L₁ = 0, L₂ = 0 be the equations of two straight lines. If P(x₁, y₁) is a point on L₁ then it satisfies the equation L₁ = 0. Similarly, if P(x₁, y₁) is a point on L₂ = 0 then it satisfies the equation.

If P(x₁, y₁) lies on L₁ or L₂, then P(x₁,y₁) satisfies the equation L₁L₂= 0.
L₁L₂ = 0 represents the pair of straight lines L₁ = 0 and L₂ = 0 and the joint equation of L₁ = 0 and L₂ = 0 is given by L₁ L₂= 0. ……………(1)
On expanding equation (1) we get and equation of the form ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 which is a second degree (non – homogeneous) equation in x and y.
Definition: If a, b, h are not all zero,then ax² + 2hxy + by² = 0 is the general form of a second degree homogeneous equation in x and y.
Definition: If a, b, h are not all zer, then ax² + 2hxy + by² + 2gx + 2 fy + c = 0 is the general form of a second degree non – homogeneous equation in x and y.

Theorem:
If a, b, h are not all zero and h² ≥ ab then ax² + 2hxy + by² = 0 represents a pair of straight lines passing through the origin.
Proof:
Case (i) : Suppose a = 0.
Given equation ax² + 2hxy + by² = 0 reduces to 2hxy + by² = 0 ^ y(2hx + by) = 0 .
Given equation represents two straight lines y = 0 ………..(1) and 2hx + by = 0 ………(2) which pass through the origin.
Case (ii): Suppose a ≠ 0.
Given equation ax² + 2hxy + by² = 0
⇒ a²x² + 2ahxy + aby² = 0
⇒ (ax)² + 2(ax)(hy) + (hy)² – (h² – ab)y² = 0
⇒ (ax + hy)² – (y\(\sqrt{h^{2}-a b}\))² = 0
[ax + y (h + \(\sqrt{h^{2}-a b}\))][ax + y (h – \(\sqrt{h^{2}-a b}\)] = 0

∴ Given equation represents the two lines
ax + hy + y\(\sqrt{h^{2}-a b}\) = 0, ax + hy – y\(\sqrt{h^{2}-a b}\) = 0 which pass through the origin.

Note 1:

• If h² > ab , the two lines are distinct.
• If h² = ab , the two lines are coincident.
• If h² < ab , the two lines are not real but intersect at a real point (the origin).
• If the two lines represented by ax² + 2hxy + by² = 0 are taken as l₁x + m₁y = 0 and l₂x + m₂y = 0 then
  ax² + 2kxy + by² = (4 x + m₁y) (x + m₂y) = 2 x² + (l₁m₂ + l₂m₁) xy + m₁m₂ y²
• Equating the coefficients of x², xy and y² on both sides, we get l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

Theorem:
If ax² + 2hxy + by² = 0 represent a pair of straight lines, then the sum of slopes of lines is \(\frac{-2 h}{b}\) product of the slopes is \(\frac{a}{b}\).
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ………….(1) and l₂x + m₂y = 0 ………….(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
Slopes of the lines (1) and (2) are –\(\frac{l_{1}}{m_{1}}\) and \(-\frac{l_{2}}{m_{2}}\).
sum of the slopes = \(\frac{-l_{1}}{m_{1}}+\frac{-l_{2}}{m_{2}}=-\frac{l_{1} m_{2}+l_{2} m_{1}}{m_{1} m_{2}}=-\frac{2 h}{b}\)
Product of the slopes = \(\left(\frac{-l_{1}}{m_{1}}\right)\left(-\frac{l_{2}}{m_{2}}\right)=\frac{l_{1} l_{2}}{m_{1} m_{2}}=\frac{a}{b}\)

Angle Between A Pair of Lines:
Theorem :
If θ is the angle between the lines represented by ax² + 2hxy + by² = 0, then cos θ = ±\(\frac{a+b}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁ x + m₁ y = 0 ………..(1) and l₂x + m₂ y = 0 …………..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

Let θ be the angle between the lines (1) and (2). Then cos θ = ±\(\frac{l_{1} l_{2}+m_{1} m_{2}}{\sqrt{\left(l_{1}^{2}+m_{1}^{2}\right)\left(l_{2}^{2}+m_{2}^{2}\right)}}\)

Note 1:
If θ is the accute angle between the lines ax² + 2hxy + by² = 0 then cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
If θ is the accute angle between the lines ax² + 2hxy + by² = 0 then tan θ = ±\(\frac{2 \sqrt{h^{2}-a b}}{a+b}\) and sin θ = \(\frac{2 \sqrt{h^{2}-a b}}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

Conditions For Perpendicular And Coincident Lines:

• If the lines ax² + 2hxy + by² = 0 are perpendicular to each other then θ = π/ 2 and cos θ = 0 ⇒ a + b = 0 i.e., co-efficient of x² + coefficient of y² = 0.
• If the two lines are parallel to each other then 0 = 0.
  ⇒ The two lines are coincident ⇒ h² = ab

Bisectors of Angles:
Theorem:
The equations of bisectors of angles between the lines a₁ x + b₁ y + c₁ = 0, a₂ x + b₂ y + c₂ = 0 are \(\frac{a_{1} x+b_{1} y+c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}\) = ±\(\frac{a_{2} x+b_{2} y+c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\)

Pair of Bisectors of Angles:
The equation to the pair bisectors of the angle between the pair of lines ax² + 2hxy + by² = 0 is h(x² – y²) = (a – b)xy (or) \(\frac{x^{2}-y^{2}}{a-b}=\frac{x y}{h}\).
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ……….(1)
and l₂x + m₂y = 0 ………(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

The equations of bisectors of angles between (1) and (2) are \(\frac{l_{1} x+m_{1} y}{\sqrt{l_{1}^{2}+m_{1}^{2}}}-\frac{l_{2} x+m_{2} y}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\) = 0 and
\(\frac{l_{1} x+m_{1} y}{\sqrt{l_{1}^{2}+m_{1}^{2}}}+\frac{l_{2} x+m_{2} y}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\) = 0

The combined equation of the bisectors is

Theorem
The equation to the pair of lines passing through (x₀, y₀) and parallel ax² + 2hxy + by² = 0
is a( x – x₀)² + 2h( x – x₀)(y – y₀) + b( y – y₀)² = 0
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ……(1) and l₂x + m₂ y = 0 …….. (2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
The equation of line parallel to (1) and passing through (x₀, y₀) is l₂(x – x₀) + m₂(y – y₀) = 0 ………(3)
The equation of line parallel to (2) and passing through (x₀, y₀) is l₂(x – x₀) + m₂(y – y₀) = 0 ………(4)

The combined equation of (3), (4) is
[l₁( x – x₀) + m₁( y – y₀)][l₂( x – x₀) + m₂(y – y₀)] = 0
⇒ l₁l₂(x – x₀)² + (l₁m₂ + l₂m₁)(x – x₀)(y – y₀) + m₁m₂(y – y₀)² = 0
⇒ a( x – x₀)² + 2h( x – x₀)( y – y₀) + b( y – y₀)2² = 0

Theorem:
The equation to the pair of lines passing through the origin and perpendicular to ax² + 2hxy + by² = 0 is bx² – 2hxy + ay² = 0 .
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ………(1) and l₂x + m₂y = 0 ……..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
The equation of the line perpendicular to (1) and passing through the origin is m₁x – l₁y = 0 ……….(3)
The equation of the line perpendicular to (2) and passing through the origin is m₂ x – l₂ y = 0 — (4)
The combined equation of (3) and (4) is
⇒ (m₁x – l₁ y)(m₂ x – l₂ y) = 0
⇒ m₁m₂x – (l₁m₂ + l₂m₁ )ny + l₁l₂ y = 0
bx² – 2hxy + ay² = 0

Theorem:
The equation to the lines passing through (x₀, y₀) and Perpendicular to ax² + 2hxy + by² = 0 is b(x – x₀)² – 2h(x – x₀)(y – y₀) + a(y – y₀)² = 0

Area of the triangle:
Theorem:
The area of triangle formed by the lines ax² + 2hxy + by² = 0 and lx + my + n = 0 is \(\frac{n^{2} \sqrt{h^{2}-a b}}{\left|a m^{2}-2 h \ell m+b \ell^{2}\right|}\)
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ………(1) and l₂x + m₂y = 0 ……..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

The given straight line is lx + my + n = 0 ………(3)
Clearly (1) and (2) intersect at the origin.
Let A be the point of intersection of (1) and (3). Then

Theorem:
The product of the perpendiculars from (α, β) to the pair of lines ax² + 2hxy + by² = 0 is \(\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 — (1) and l₂x + m₂ y = 0 …………..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
The lengths of perpendiculars from (α, β) to
the line (1) is p = \(\frac{\left|l_{1} \alpha+m_{1} \beta\right|}{\sqrt{l_{1}^{2}+m_{1}^{2}}}\)
and to the line (2) is q = \(\frac{\left|l_{2} \alpha+m_{2} \beta\right|}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\)

∴ The product of perpendiculars is

Pair of Lines-Second Degree General Equation:
Theorem:
If the equation S ≡ ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 represents a pair of straight lines then
(i) Δ ≡ abc + 2fgh – af² – bg² – ch² =0 and
(ii) h² ≥ ab, g² ≥ ac, f² ≥ bc
Proof:
Let the equation S = 0 represent the two lines l₁x + m₁ y + n₁ = 0 and l₂ x + m₂ y + n₂ = 0. Then
ax2 + 2hxy + by2 + 2 gx + 2 fy + c
≡ (l₁x + m₁y + n₁)(l₂ x + m₂ y + n₂) = 0

Equating the co-efficients of like terms, we get
l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b, and l₁n₂ + l₂n₁ = 2g , m₁n₂ + m₂n₁ = 2 f , n₁n₂ = c

(i) Consider the product(2h)(2g)(2f)
= (l₁m₂ + l₂m₁)(l₁n₂ + l₂n₁)(m₁n₂ + m₂n₁)
= l₁l₂ (m₁²n₂ + m₂²n₁²) + m₁m₂ (l₁²n₂² + l₂²n₁²) + n₁n₂ (l₁²m₂² + l₂²m₁²) + 2l₁l₂m₁m₂n₁n₂
= l₁l₂[(m₁n₂ + m₂n₁) – 2m₁m₂n₁n₂] + m₁m₂[(l₁n₂ + l₂n₁) – 2l₁l₂n₁n₂] + n₁n₂[(l₁m₂ + l₂m₁) – 2l₁l₂m₁m₂] + 2l₁l₂m₁m₂n₁n₂
= a(4 f² – 2bc) + b(4g² – 2ac) + c(4h² – 2ab)
8 fgh = 4[af² + bg² + ch² – abc]
abc + 2 fgh – af² – bg² – ch² = 0

(ii) h² – ab = \(\left(\frac{l_{1} m_{2}+l_{2} m_{1}}{2}\right)^{2}\) – l1l2m1m2 = \(\frac{\left(l_{1} m_{2}+l_{2} m_{1}\right)^{2}-4-l_{1} l_{2} m_{1} m_{2}}{4}\)
= \(\frac{\left(l_{1} m_{2}-l_{2} m_{1}\right)^{2}}{4}\) ≥ 0
Similarly we can prove g² > ac and f² ≥ bc

Note :
If A = abc + 2 fgh – af² – bg² – ch² = 0 , h² ≥ ab, g² ≥ ac and f² ≥ bc, then the equation S ≡ ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 represents a pair of straight lines

Conditions For Parallel Lines-Distance Between Them:
Theorem:
If S = ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 represents a pair of parallel lines then h² = ab and bg² = af². Also the distance between the two parallel lines is 2\(\sqrt{\frac{g^{2}-a c}{a(a+b)}}\) (or) 2\(\sqrt{\frac{f^{2}-b c}{b(a+b)}}\)
Proof:
Let the parallel lines represented by S = 0 be
lx + my + n₁ = 0 ……….(1) lx + my + n₂ = 0 ………..(2)
ax² + 2hxy + 2gx + 2 fy + c
= (lx + my + n₁)(lx + my + n₂)

Equating the like terms
l² = a ………(3)
2lm = 2h …………(4)
m² = b ………..(5)
l(n₁ + n₂) = 2g …….(6)
m(n₁ + n₂) = 2 f ….(7)
n₁n₂ = c …….(8)

From (3) and (5), l²m² = ab and from (4) h² = ab .

Point of Intersection of Pair of Lines:
Theorem:
The point of intersection of the pair of lines represented by
a² + 2hxy + by² + 2gx + 2fy + c = 0 when h² > ab is \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)
Proof:
Let the point of intersection of the given pair of lines be (x₁, y₁).
Transfer the origin to (x₁, y₁) without changing the direction of the axes.
Let (X, Y) represent the new coordinates of (x, y). Then x = X + x₁ and y = Y + y₁.
Now the given equation referred to new axes will be
a( X + x₁)2 + 2h( X + x₁)(Y + y₁) +b(Y + y₁)2 + 2 g (X + x₁) + 2 f (Y + y₁) + c = 0
⇒ aX² + 2hXY + bY² + 2 X (ax₁ + hy₁ + g) + 2Y(hx₁ + by₁ + f) +(ax₁² + 2hx₁y₁ + by² + 2 gx₁ + 2 fy₁ + c) = 0

Since this equation represents a pair of lines passing through the origin it should be a homogeneous second degree equation in X and Y. Hence the first degree terms and the constant term must be zero. Therefore,
ax₁ + hy₁ + g = 0
hx₁ + by₁ + f = 0
ax₁² + 2hx₁ y₁ + by₁² + 2 gx₁ + 2 fyx + c = 0
But (3) can be rearranged as
x₁(ax₁ + hy + g) + y (hx₁ + byx + f) + (gx₁ + fq + c) = 0
⇒ gx₁ + fy₁ + c = 0 ………..(4)

Solving (1) and (2) for x₁ and y₁

Hence the point of intersection of the given pair of lines is \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

Theorem:
If the pair of lines ax² + 2hxy + by² = 0 and the pair of lines ax² + 2hxy + by² + 2gx + 2fy + c = 0 form a rhombus then (a – b) fg+h(f² – g²) = 0.
Proof:
The pair of lines ax² + 2hxy + by² = 0 …………(1) is parallel to the lines ax² + 2hxy + by² + 2gx + 2fy + c = 0 ……….. (2)

Now the equation
ax² + 2hxy + by² + 2gx + 2 fy + c + λ(ax² + 2hxy + by²) = 0

Represents a curve passing through the points of intersection of (1) and (2).
Substituting λ = -1, in (3) we obtain 2gx + 2fy + c = 0 …(4)
Equation (4) is a straight line passing through A and B and it is the diagonal \(\overline{A B}\)

The point of intersection of (2) is C = \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)
⇒ Slope of \(\overline{O C}=\frac{g h-a f}{h f-b g}\)
In a rhombus the diagonals are perpendicular ⇒ (Slope of \(\overline{O C}\)) (Slope of \(\overline{A B}\)) = -1
⇒ \(\left(\frac{g h-a f}{h f-b g}\right)\left(-\frac{g}{f}\right)\) = -1
⇒ g²h – afg = hf² – bfg
⇒ (a – b)fg + h(f² – g²) = 0
\(\frac{g^{2}-f^{2}}{a-b}=\frac{f g}{h}\)

Theorem:
If ax² + 2hxy + by² = 0 be two sides of a parallelogram and px + qy = 1 is one diagonal, then the other diagonal is y(bp – hq) = x(aq – hp)
proof:
Let P(x₁, y₁) and Q(x₂, y₂) be the points where the digonal

px + qy = 1 meets the pair of lines.
\(\overline{O R}\) and \(\overline{P Q}\) biset each other at M(α, β)
∴ α = \(\frac{x_{1}+x_{2}}{2}\) and β = \(\frac{y_{1}+y_{2}}{2}\)

Eliminating y from ax² + 2hxy+by² = 0
and px + qy = 1 ………..(2)
ax² + 2hx\(\left(\frac{1-p x}{q}\right)\) + b\(\left(\frac{1-p x}{q}\right)^{2}\) = 0
⇒ x² (aq² – 2hpq + bp²) + 2 x(hp – bp) + b = 0

The roots of this quadratic equation are x₁ and x₂ where
x₁ + x₂ = \(-\frac{2(h q-b p)}{a q^{2}-2 h p q-b p^{2}}\)
⇒ α = \(\frac{(b p-h q)}{\left(a q^{2}-2 h p q+b p^{2}\right)}\)

Similarly, by eliminating x from (1) and (2) a quadratic equation in y is obtained and y₁,
y₂ are its roots where
y₁ + y₂ = \(-\frac{2(h p-a q)}{a q^{2}-2 h p q-n p^{2}}\) ⇒ β = \(\frac{(a q-h p)}{\left(a q^{2}-2 h p q+b p^{2}\right)}\)
Now the equation to the join of O(0, 0) and M(α, β) is (y – 0)(0 – α) = (x – 0)(0 – β)
⇒ αy = βx
Substituting the values of α and β, the equation of the diagonal OR
is y(bp – hq) = x(aq – hp).

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Addition of Vectors Solutions Exercise 4(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Addition of Vectors Solutions Exercise 4(a)

I.

Question 1.
ABCD is a Parallelogram. If L and M are the middle points of BC and CD, respectively, then find (i) AL and AM in terms of AB and AD (ii) λ, if AM = λ AD – LM.
Solution:

Question 2.
In ∆ABC, P, Q, and R are the midpoints of the sides AB, BC, and CA respectively. If D is any point.
(i) then express \(\overline{\mathrm{DA}}+\overline{\mathrm{DB}}+\overline{\mathrm{DC}}\) interms of \(\overline{D P}\), \(\overline{D Q}\) and \(\overline{D R}\).
(ii) If \(\overline{\mathbf{P A}}+\overline{\mathbf{Q B}}+\overline{\mathbf{R C}}=\bar{\alpha}\) then find \(\bar{\alpha}\)
Solution:

Question 3.
Let \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+2 \overline{\mathbf{j}}+3 \overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}=\mathbf{3} \overline{\mathbf{i}}+\overline{\mathbf{j}}\). Find the unit vector in the direction of \(\overline{\mathbf{a}}+\overline{\mathbf{b}}\).
Solution:

Question 4.
If the vectors \(-3 \overline{\mathbf{i}}+4 \bar{j}+\lambda \overline{\mathbf{k}}\) and \(\mu \bar{i}+8 \bar{i}+6 \bar{k}\) are coilinear vectors , then find λ and µ.
Solution:

Question 5.
ABCDE is a pentagon. If the sum of the vectors \(\overline{\mathrm{AB}}, \overline{\mathrm{AE}}, \overline{\mathrm{BC}}, \overline{\mathrm{DC}}, \overline{\mathrm{ED}}\) and \(\overline{\mathbf{A C}}\) is λ \(\overline{\mathbf{A C}}\), then find the value of λ.
Solution:

Question 6.
If the position vectors of the points A, B and C are \(-2 \overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}},-4 \overline{\mathbf{i}}+2 \overline{\mathbf{j}}+2 \overline{\mathbf{k}}\) and \(6 \bar{i}-3 \bar{j}-13 \bar{k}\) respectively and \(\overline{\mathbf{A B}}=\lambda \overline{\mathrm{AC}}\), then find the value of λ.
Solution:

Question 7.
If \(\overline{\mathrm{OA}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathrm{AB}}=3 \bar{i}-2 \overline{\mathbf{j}}+\overline{\mathbf{k}}\), \(\overline{B C}=\bar{i}+2 \bar{j}-2 \bar{k}\) and \(\overline{C D}=2 \bar{i}+\bar{j}+3 \bar{k}\), then find the vector \(\overline{O D}\).
Solution:

Question 8.
\(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+5 \overline{\mathbf{j}}+\overline{\mathbf{k}}\) and \(\bar{b}=4 \bar{i}+m \bar{j}+n \bar{k}\) are collinear vectors, then find m and n.
Solution:

Question 9.
Let \(\bar{a}=2 \bar{i}+4 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}, \bar{b}=\hat{i}+\bar{j}+\bar{k}\) and \(\bar{c}=\bar{j}+2 \bar{k}\). Find the unit vector in the opposite direction of \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}\).
Solution:

Question 10.
Is the triangle formed by the vectors \(3 \bar{i}+5 \bar{j}+2 \bar{k}, 2 \bar{i}-3 \bar{j}-5 \bar{k}\) and \(-5 \bar{i}-2 \bar{j}+3 \bar{k}\) equilateral?
Solution:

Question 11.
If α, β and γ be the angles made by the vector \(3 \bar{i}-6 \bar{i}+2 \bar{k}\) with the positive directions of the co-ordinate axes, then find cos α, cos β, cos γ.
Solution:
Unit vectors along the co-ordinate axes are respectively \(\bar{i}, \bar{j}, \bar{k}\).

Question 12.
Find the angles made by the straight line passing through the points (1, -3, 2) and (3, -5, 1) with the co-ordinate axes.
Solution:
Unit vectors along the co-ordinate axes are respectively \(\bar{i}, \bar{j}, \bar{k}\).
Let A(1, -3, 2) and B(3, -5, 1) be two given points.
Let ‘O’ be the origin. Then

II.

Question 1.
If \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}=\alpha \overline{\mathbf{d}}, \overline{\mathbf{b}}+\overline{\mathbf{c}}+\overline{\mathbf{d}}=\beta \overline{\mathbf{a}}\) and \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathrm{c}}\) are non-coplanar vectors, then show that \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}+\overline{\mathbf{d}}=\mathbf{0}\).
Solution:

Question 2.
\(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are non-coplanar vectors. Prove that the following four points are coplanar.
(i) \(-\overline{\mathbf{a}}+4 \overline{\mathbf{b}}-3 \bar{c}, \quad 3 \bar{a}+2 \bar{b}-5 \bar{c}\), \(-3 \overline{\mathbf{a}}+8 \overline{\mathbf{b}}-5 \overline{\mathbf{c}},-3 \overline{\mathbf{a}}+2 \overline{\mathbf{b}}+\overline{\mathbf{c}}\)
Solution:
Let ‘O’ be the origin and A, B, C, D be the four points.


4 + 2x + 2y = 0 ……..(1)
-2 – 4x + 2y = 0 ……..(2)
-2 + 2x – 4y = 0 …….(3)
Solve (1) and (3)
6y + 6 = 0 ⇒ y = -1
Substitute in (1)
2x + 2 (-1) + 4 = 0
⇒ 2x + 2 = 0
⇒ x = -1
Substitute x = -1, y = -1 in (2)
-2 – 4(-1) + 2(-1) = -4 + 4 = 0
∴ The vectors \(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}, \overline{\mathrm{AD}}\) are coplanar
⇒ A, B, C, D are coplanar
Hence the given points are coplanar.

(ii) \(6 \bar{a}+2 \bar{b}-\bar{c}, 2 \bar{a}-\bar{b}+3 \bar{c},-\bar{a}+2 \bar{b}-4 \bar{c},\)\(-12 \bar{a}-\bar{b}-3 \bar{c}\)
Solution:
Let O be the origin. Let A, B, C, D be the given points.

Let us suppose that one vector can be expressed as a linear combination of the other two.

∵ \(\bar{a}, \bar{b}, \bar{c}\) are non-coplanar vectors.
7x + 18y – 4 = 0 ………(1)
-3 + 3y = 0 ⇒ y = 1 …….(2)
3x + 2y + 4 = 0 ………(3)
Substitute y =1 in (3)
3x + 2 + 4 = 0 ⇒ x = -2
Substitute x = -2 and y = 1 in (1)
7(-2) + 18(1) – 4 = 0 ⇒ 0 = 0
Hence \(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}\) and \(\overline{\mathrm{AD}}\) are coplanar.
⇒ The points A, B, C, D are coplanar.

Question 3.
If \(\overline{\mathbf{i}}, \overline{\mathbf{j}}, \overline{\mathbf{k}}\) are unit vectors along the positive directions of the coordinate axes, then show that the four points \(4 \overline{\mathbf{i}}+5 \overline{\mathbf{j}}+\overline{\mathbf{k}},-\overline{\mathbf{j}}-\overline{\mathbf{k}}, 3 \overline{\mathbf{i}}+9 \overline{\mathbf{j}}+4 \overline{\mathbf{k}}\) and \(-4 \bar{i}+4 \bar{j}+4 \bar{k}\) are coplanar.
Solution:
Let ‘O’ be the origin and let A, B, C, D be the given points.


⇒ The given points A, B, C, D are coplanar.
Second Method:
\(\left[\begin{array}{lll}
\overline{\mathrm{AB}} & \overline{\mathrm{AC}} & \overline{\mathrm{AD}}
\end{array}\right]\) = \(\left|\begin{array}{ccc}
-4 & -6 & -2 \\
-1 & 4 & 3 \\
-8 & -1 & 3
\end{array}\right|\)
= -4(12 + 3) + 6(-3 + 24) – 2(1 + 32)
= -60 + 126 – 66
= 0
Hence the vectors \(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}\) and \(\overline{\mathrm{AD}}\) are coplanar.
⇒ The given points A, B, C, D are coplanar.

Question 4.
If a, b, c are non-coplanar vectors, then test for the collinearity of the following points whose position vectors are given by
(i) \(\bar{a}-2 \bar{b}+3 \bar{c}, 2 \bar{a}+3 \bar{b}-4 \bar{c},-7 \bar{b}+10 \bar{c}\)
Solution:
Let ‘O’ be the origin. A, B, C be the given points.
Then \(\overline{\mathrm{OA}}=\overline{\mathrm{a}}-2 \overline{\mathrm{b}}+3 \overline{\mathrm{c}}\)

(ii) \(3 \bar{a}-4 \bar{b}+3 \bar{c}\), \(-4 \bar{a}+5 \bar{b}-6 \bar{c}\), \(4 \overline{\mathbf{a}}-7 \overline{\mathbf{b}}+6 \overline{\mathbf{c}}\)
Solution:

(iii) \(\begin{aligned}
&2 \bar{a}+5 \bar{b}-4 \bar{c}, \bar{a}+4 \bar{b}-3 \bar{c}, \\
&4 \bar{a}+7 \bar{b}-6 \bar{c}
\end{aligned}\)
Solution:
Let ‘O’ be the origin and A, B, C be the given points.
Then \(\overline{\mathrm{OA}}=2 \overline{\mathrm{a}}+5 \overline{\mathrm{b}}-4 \overline{\mathrm{c}}\), \(\overline{\mathrm{OB}}=\overline{\mathrm{a}}+4 \overline{\mathrm{b}}-3 \overline{\mathrm{c}}\)

∴ The points A, B, C are collinear.

III.

Question 1.
In the Cartesian plane, O is the origin of the coordinate axes. A person starts at O and walks a distance of 3 units in the NORTH-EAST direction and reaches point P. From P he walks 4 units of distance parallel to NORTH-WEST direction and reaches the point Q. Express the vector \(\overline{\mathbf{O Q}}\) in terms of \(\overline{\mathbf{i}}\) and \(\overline{\mathbf{j}}\) (observe that ∠XOP = 45°)
Solution:
‘O’ the origin of co-ordinate axes.

Question 2.
The points O, A, B, X and Y are such that \(\overline{\mathbf{O A}}=\overline{\mathbf{a}}, \overline{\mathbf{O B}}=\overline{\mathbf{b}}, \overline{\mathbf{O X}}=\mathbf{3} \overline{\mathbf{a}}\) and \(\overline{\mathbf{O Y}}=\mathbf{3} \overline{\mathbf{b}}\). Find \(\overline{\mathbf{B X}}\) and \(\overline{\mathbf{A Y}}\) interms of \(\bar{a}\) and \(\bar{b}\). Futher, if the point P divides AY in the ratio 1 : 3, then express \(\overline{\mathrm{BP}}\) interms of \(\bar{a}\) and \(\bar{a}\).
Solution:

Question 3.
If ∆OAB, E is the midpoint of AB and F is a point on OA such that OF = 2(FA). If C is the point of intersection of \(\overline{\mathrm{OE}}\) and \(\overline{\mathrm{BF}}\), then find the ratios OC : CE and BC : CF.
Solution:

Question 4.
Point E divides the segment PQ internally in the ratio 1 : 2 and R is any point not on the line PQ. If F is a point on QR such that QF : FR = 2 : 1, then show that EF is parallel to PR.
Solution:

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Pair of Straight Lines Solutions Exercise 4(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1B Pair of Straight Lines Solutions Exercise 4(a)

I.

Question 1.
Find the acute angle between the pair of lines represented by the following equations.
(i) x² – 7xy + 12y² = 0
(ii) y² – xy – 6x² = 0
(iii) (x cos α – y sin α)² = (x² + y²) sin² α
(iv) x² + 2xy cot α – y² = 0
Solution:
(i) x² – 7xy + 12y² = 0
a = 1, b = 12, h = –\(\frac{7}{2}\)

(ii) y² – xy – x² = 0
a = -6, b = 1, h = –\(\frac{1}{2}\)
\(\cos \theta=\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

(iii) (x cos α – y sin α)² = (x² + y²) sin² α
x2 cos² α + y² sin² a – 2xy cos α sin α = x² sin² α + y² sin² α
∴ x² (cos² α – sin² α) – 2xy cos α sin α = 0
x².cos 2α – xy sin 2α = 0
a = cos 2α, b = 0, 2h = -sin 2α
\(\cos \theta=\frac{\|\cos 2 \alpha+0\|}{\sqrt{(\cos 2 \alpha-0)^{2}+\sin ^{2} 2 \alpha}}\)
= cos 2α
∴ θ = 2α

(iv) x² + 2xy cot a – y² = 0
a + b = 1 – 1 = 0
∴ θ = \(\frac{\pi}{2}\)

II.

Question 1.
Show that the following pairs of straight lines have the same set of angular bisectors (that is they are equally inclined to each other).
i) 2x² + 6xy + y² = 0,
4x² + 18xy + y² = 0.
ii) a²x² + 2h(a + b) xy + b²y² = 0,
ax² + 2hxy + by² = 0, a + b ≠ 0.
iii) ax² + 2hxy + by² + λ(x² + y²) = 0; (λ ∈ R),
ax² + 2hxy + by² = 0.

Solution:
(i) Combined equation of OA, OB is
2x² + 6xy + y² = 0
Equation of the pair of bisectors is
3(x² – y²) = (2 – 1) xy
3(x² – y2² = xy ………… (1)
Combined equation of OP, OQ is
4x² + 18xy + y² = 0
Equation of the pair of bisectors is
9(x² – y²) = (4 – 1) xy 9(x² – y²) = 3xy
3(x² – y²) = xy ………….. (2)
(1), (2) are same.
∴ OA, OB and OP, OQ are inclined to each other.

(ii) Combined equation of OA, OB is
a²x² + 2h(a + b) xy + b²y² = 0
Equation of the pair of bisectors is
h (a + b) (x² – y²) = (a² – b²) xy
h (a + b) (x² – y²) = (a + b)(a – b) xy
i.e., h(x² – y²) = (a – b) xy ………… (1)
Combined equation of OP, OQ is
ax² + 2hxy + by² = 0
Equation of the pair of bisectors is
h (x² – y²) = (a – b) xy …………. (2)
(1), (2) are same.
∴ OA, OB and OP, OQ are equally inclined to each other.

(iii) Combined equation of OA, OB is
ax² + 2hxy + by² + λ(x² + y²) = 0
(a + λ) x² + 2hxy + (b + λ) y² = 0
Equation of the pair of bisectors of OA, OB is
h (x2 – y2) = (a + λ – b – λ)xy
= (a – b)xy ……….. (1)
Combined equation of OP, OQ is
ax² + 2hxy + by² = 0
Equation of the pair of bisectors of OP, OQ is
h(x² – y²) = (a – b) xy …………. (2)
(1), (2) are same.
∴ OA, OB and OP, OQ are equally inclined to each other.

Question 2.
Find the value of h, if the slopes of the lines represented by 6x² + 2hxy + y² = 0 are in the ratio 1: 2.
Solution:
Combined equation of the lines is
6x² + 2hxy + y² = 0
Suppose individual equations of the given lines are y = m₁x and y = m₂x

Question 3.
If ax² + 2hxy + by² = 0 represents two straight lines such that the slope of one line is twice the slope of the other, prove that 8h² = 9ab.
Solution:
Combined equation of the lines is
ax² + 2hxy + by² =0
Suppose, y = m₁x and y = m₂x are the individual equations of the lines.
∴ m₁ + m₂ = –\(\frac{2h}{b}\), m₁m₂ = \(\frac{a}{b}\)
Given m₂ = 2m₁
∴ 3m₁ = –\(\frac{2h}{b}\) ; 2m₁² = \(\frac{a}{b}\)
m₁ = –\(\frac{2h}{3}\) ; m₁² = \(\frac{a}{2b}\)
∴ (-\(\frac{2h}{3b}\))² = \(\frac{a}{2b}\)
\(\frac{4 h^{2}}{9 b^{2}}=\frac{a}{2 b}\)
8h² = 9ab.

Question 4.
Show that the equation of the pair of straight lines passing through the origin and making an angle of 30° with the line 3x – y – 1 = 0 is 13x² + 12xy – 3y² = 0

Solution:
Equation of AB is 3x – y – 1 = 0
OA, OB make an angle of 30° with AB and pass through the origin.
Suppose slope of OA is m
∴ Equation of OA is
y – 0 = m (x – 0) = mx or mx – y = 0

∴ \(\frac{\sqrt{3}}{2}=\frac{|3 m+1|}{\sqrt{10} \sqrt{m^{2}+1}}\)
Squaring and cross multiplying
\(\frac{3\left(m^{2}+1\right)}{4}=\frac{(3 m+1)^{2}}{10}\)
15(m² + 1) = 2 (3m + 1)²
15m² + 15 = 2 (9m² + 6m + 1)
= 18m² + 12m + 2
3m² + 12m -13 = 0
Suppose m₁, m₂ are two roots of the equation
m₁ + m₂ = -4, m₁ m₂ = \(\frac{-13}{3}\)
Combined equation of OA and OB is
(m₁x – y) (m₂x – y) = 0
m₁m₂x² – (m₁ + m₂) xy + y² = 0
\(\frac{-13}{3}\)x² + 4xy + y² = 0
-13 x² + 12xy + 3y² = 0 or
13x² – 12xy – 3y² = 0

Question 5.
Find the equation to the pair of straight lines passing through the origin and making an acute angle a with the straight line x + y + 5 = 0.

Solution:
Equation of AB is x + y + 5 = 0 ……….. (1)
Slope of AB = – λ
Suppose OA and OB are the required lines
Suppose equation of OA is
y = mx ⇒ mx – y = 0

2(m² + 1) cos² α = (m – 1)²
2(m² + 1) = \(\frac{(m-1)^{2}}{\cos ^{2} \alpha}\) = (m – 1)² sec² α.
2m² + 2 = m² sec² α – 2m sec² α + sec² α.
m² (sec² α – 2) – 2m sec² α + (sec² α – 2) = 0
m₁ + m₂ = \(\frac{2 \sec ^{2} \alpha}{\sec ^{2} \alpha-2}\), m₁m₂ = 1
Combined equation of OA and OB is
(y – m₁x) (y – m₂x) = 0
y² (m₁ + m₂) xy + m₁m₂ x² = 0

Combined equation of OA and OB is
x² + 2xy sec 2a + y² = 0

Question 6.
Show that the straight lines represented by (x + 2a)² – 3y² = 0 and x = a form an equilateral triangle.
Solution:
Combined equation of OA, OB is
(x + 2a)² – 3y² = 0
(x + 2a)² – (√3y)² =0
(x + 2a + √3 y) (x + 2a – √3 y) = 0

Equation of OA is
x + √3y + 2a = 0 ………….. (1)
Equation of OB is
x – √3y + 2a = 0 ………. (2)
Equation of AB is x – a = 0

∴ ∠OBA =60°
∴ ∠AOB = 180°- (∠OAB + ∠OBA)
= 180° – (60° + 60°)
= 180° – 120°
= 60°
∴ ∆OAB is an equilateral triangle.

Question 7.
Show that the pair of bisectors of the angles between the straight lines (ax + by)² = c (bx – ay)², c > 0 are parallel and perpen-dicular to the line ax+ by + k= 0.
Solution:
Combined equation of the given lines is (ax + by)² = c (bx – ay)²
a²x² + b²y² + 2ab xy = c (b²x² + a²y² – 2abxy)
= cb²x² +ca²y² – 2cabxy
(a² – cb²)x² + 2ab (1 + c²) xy + (b² – ca²)y² = 0
Equation of the pair of bisectors is
h (x – y²) = (a – h) xy ^
ab (1 + c) (x² – y²)
= (a² – cb² – b² + ca²) (x² – y²) = 0
= (a² – b²)(1 + c) xy.
i.e., ab (x² – y²) – (a² – b²) xy = 0
(ax + by) (bx – ay) = abx² – a2xy +b²xy – aby²
= ab(x² – y²) – (a² – b²)xy
∴ The equation of the pair of bisectors are (ax + by) (bx – ay) = 0
The bisectors are ax + by = 0 and bx – ay = 0
ax + by = 0 is parallel to ax + by + k = 0
bx – ay = 0 is perpendicular to ax + by +k=0.

Question 8.
The adjacent sides of a parallelogram are 2x² – 5xy + 3y² = 0 and one diagonal is x + y + 2 = 0. Find the vertices and the other diagonal.

Solution:
Combined equation of OA and OB is 2x² – 5xy + 3y² = 0
Equation of AB is x + y+ 2 = 0
y = -(x + 2)
Substituting in (1)
2x² + 5x (x + 2) + 3(x +2)² = 0
2x² + 5x² + 10x + 3(x² + 4x + 4) = 0
7x² + 10x + 3x² + 12x + 12 = 0
10x² + 22x + 12 = 0
5x² + 11x + 6 = 0
(x + 1) (5x + 6) = 0
x + 1 = 0 or 5x + 6 = 0
x = -1 or 5x = -6
x = –\(\frac{6}{5}\)
y = -(x + 2)
x = -1 ⇒ y = – (-1 + 2) = – 1
⇒ co-ordinates of A are (-1, -1)
x = –\(\frac{6}{5}\) ⇒ y = -(\(\frac{6}{5}\)+ 2) = –\(\frac{4}{5}\)
⇒ co-ordinates of B are (-\(\frac{6}{5}\), –\(\frac{4}{5}\))
Suppose the diagonals AB, OC intersect in O’ O’ bisects AB and OC.
Suppose co-ordinates of C are (x, y)
Midpoint of OC = Midpoint of AB

∴ The vertices are O(0, 0), A(-1, -1)
C(-\(\frac{11}{5}\), –\(\frac{9}{5}\)), B(-\(\frac{6}{5}\), –\(\frac{4}{5}\))

Question 9.
Find the centroid and the area of the triangle formed by the following lines.
(i) 2y² – xy – 6x² = 0, x + y + 4 = 0
(ii) 3x² – 4xy + y² = 0, 2x – y = 6

Solution:
(i) Combined equation of OA, OB is
2y² – xy – 6x² = 0 ………… (1)
Equation of AB is x + y + 4 = 0
y = – (x + 4) ………… (2)
Substituting in (1)
2(x + 4)² + x (x + 4) – 6x² = 0
2(x² + 8x + 16) + x² + 4x – 6x² = 0
2x² + 16x + 32 + x² + 4x – 6×2 = 0
-3x² + 20x + 32 = 0
3x² – 20x – 32 = 0
(3x + 4) (x-8) = 0
3x + 4 = 0 or x – 8 = 0
x = –\(\frac{4}{3}\) or 8

Case (i) : x = –\(\frac{4}{3}\)
y = – (x + 4)
= -(\(\frac{-4}{3}\) + 4) = –\(\frac{8}{3}\)
Co – ordinates of A are (-\(\frac{4}{3}\), –\(\frac{8}{3}\))

Case (ii) : x = 8
y = -(x + 4) = – (8 + 4) = – 12
Co-ordinates of B are (8, – 12)
Suppose G is the centroid of ∆ AOB
Co-ordinates of G are

(ii) Combined equation of OA, OB is
3x² – 4xy + y² = 0 ……………. (1)
Equation of AB is 2x – y = 6
y = 2x – 6 ……………. (2)
Substituting in (1)
3x² – 4x (2x – 6) + (2x – 6)² = 0
3x² + 8x² + 24x + 4x² + 36 – 24x = 0
– x + 36 = 0
x² – 36 = 0
(x + 6) (x – 6) = 0
x + 6 = 0 or x – 6 = 0
x = – 6 or 6
y = 2x – 6
x = 6 ⇒ y = 12 – 6 = 6
Go -ordinates of A are (6, 6)
x = -6 ⇒ y = – 12 – 6 = -18
Co-ordinates of B are (-6, -18)
Co-ordinates of G ate
(\(\frac{0+6-6}{3}\), \(\frac{0+6-18}{3}\)) = (0, -4)
∆OAB = \(\frac{1}{2}\)|x₁y₂ – x₂y₁|
= \(\frac{1}{2}\)|16 (-18) – (-6). 6|
= \(\frac{1}{2}\)|- 108 + 36|
= \(\frac{1}{2}\) . 72 =36 sq.units

Question 10.
Find the equation of the pair of lines intersecting at (2, -1) and
(i) perpendicular to the pair
6x² – 13xy – 5y² = 0 and
(ii) parallel to the pair
6x² – 13xy – 5y² = 0.
Solution:
Equation of OA, OB is 6x² – 13xy – 5y² = 0
(i) Equation of the pair of lines through (x₁ y₁) and perpendicular to
ax² + 2hxy + by² = 0 is
b(x – x₁)² – 2h(x – x₁) (y – y₁) + a (y – y₁)² = 0
Equation of the perpendicular pair of lines is
-5(x – 2)² + 13(x – 2) (y + 1) + 6(y+ 1)² =0
-5(x² – 4x + 4) + 13(xy + x – 2y – 2) + 6(y² + 2y + 1) = 0
-5x² + 20x – 20 + 13xy + 13x – 26y – 26 + 6y² + 12y + 6 = 0
-5x² + 13xy + 6y² + 33x – 14y – 40 = 0
or 5x² – 13xy – 6y² – 33x + 14y + 40 = 0

(ii) Equation of the pair of lines through (x₁, y₁) and parallel to ax² + 2hxy + by² = 0 is
a(x – x₁)² + 2h (x – x₁) (y – y₁) + b (y – y₁)² = 0
Equation of the pair of parallel lines is
6(x- 2)² – 13(x – 2) (y + 1) – 5(y + 1)² = 0
6(x² – 4x + 4) – 13(xy + x – 2y – 2) – 5(y² + 2y + 1) = 0
6x² – 24x + 24 – 13xy – 13x + 26y + 26 – 5y² – 10y – 5 = 0
6x² – 13xy – 5y² – 37x + 16y + 45 = 0.

Question 11.
Find the equation of the bisector of the acute angle between the lines
3x – 4y + 7 = 0 and 12x + 5y – 2 = 0
Solution:
Given lines
3x – 4y + 7 = 0 ………….. (1)
12x + 5y – 2 = 0 ………… (2)
The equations of bisector’s angles between (1) & (2) is

13 (3x – 4y + 7) ± 5 (12x + 5y- 2) = 0
(39x – 52y + 51) ± (60x + 25y – 10) = 0

(i) 39x- 52y + 51 + 60x + 25y- 10 = 0
99x-27y + 41 = 0 ……… (3)

(ii) (39x – 52y + 51) – (60x + 25y- 10) = 0
39x – 52y + 51 – 60x – 25y +10 = 0
– 21x – 77y + 61 =0
21x + 77y- 61 = 0 ……… (4)
Let ‘θ’ be the angle between (1), (4)

∴ (4) is obtuse angle bisector, then other one (3) is the accute angle bisector.
∴ 99x – 27y + 41 = 0 is the accute angle bisector.

Question 12.
Find the equation of the bisector of the obtuse angle between the lines x + y – 5 = 0 and x – 7y + 7 = 0
Solution:
Given lines
x + y- 5 = 0 ………. (1)
x – 7y + 7 = 0 ……..(2)
The equations of bisectors of angles between (1), (2) is

⇒ (5x + 5y-25)±(x-7y + 7) = 0

(i) 5x + 5y – 25 + x – 7y + 7 = 0
6x – 2y – 18 = 0
3x-y-9 = 0 ………. (3)

(ii) (5x + 5y – 25) – (x – 7y + 7) = 0
4x+ 12y – 32 = 0
x + 3y – 8 = 0 ……… (4)
Let ‘θ’ be the angle between (1), (4)

∴ (4) is the ocute angle bisector, then other one 3x – y – 9 = 0 is the obtuse angle bisector.

III.

Question 1.
Show that the lines represented by (lx + my)² – 3(mx – ly)² = 0 and lx + my + n = 0 form an equilateral triangle with area \(\frac{n^{2}}{\sqrt{3}\left(l^{2}+m^{2}\right)}\).
Solution:
Combined equation of A and is
(lx + my)² – 3(mx – ly)² = 0
l²x² + m²y² + 2lmxy – 3m²x² – 3l²y² + 6 lmxy = 0
(l² – 3m²) x² + 8lmxy + (m² – 3l²) y² = 0

Combined equation of the bisectors of OA
and OB is h (x² – y²) = (a – b) xy
4 lm (x² – y²) = (l² – 3m² – m² + 3l²) xy
4 lm (x² – y²) = 4(l² – m²) xy
lmx² – (l² – m²)xy – lmy² = 0
(lx – my) (mx – ly) = 0
lx + my = 0 and mx – ly = 0
∴ The bisectors mx – ly = 0 is perpendicular to AB whose equation lx + my + n = 0
OAB is an scales triangle and ∠AOB = 60°
OAB is an equalated tringle
P = Length of the X lan prove P and AB

Question 2.
Show that the straight tines represented by 3x² + 48xy + 23y² = 0 and 3x – 2y + 13 = 0 form an equilateral triangle of area \(\frac{13}{\sqrt{3}}\) sq.umts.
Solution:
Combined equation of OA, OB is
3x² + 48xy + 23y² = 0 …………. (1)
Equation of AB is 3x – 2y + 13 = 0 …….. (2)
(1) can be written as
(9x² – 12xy + 4y²) – 3(4x² + 12xy + 9y²) = 0
i.e., (3x – 2y)² – 3(2x + 3y)² = 0
⇒ [(3x – 2y) + √3(2x +3y)] [(3x – 2y) – √3(2x+3y)] = 0
⇒ [(3 + 2√3)x+ (3√3 – 2)y] [(3 – 2√3)x – (3√3 + 2)y]=0
Equation of OA is
(3 + 2√3)x – (3√3 – 2)y = 0 ………….. (1)
Equation of OB is
(3 – 2√3)x – (3√3 +2)y =0 ………… (2)

∴ OAB is an equilateral triangle.

Question 3.
Show that the equation of the pair of lines bisecting the angles between the pair of bisectors of the angles between the pair of lines ax² + 2hxy + by² = 0 is (a – b) (x² – y²) + 4hxy = 0
Solution:
Equation of the given lines is
ax² + 2hxy + by² = 0
Equation of the pair of bisectors is
h(x² – y²) = (a – b)xy …………. (1)
hx² – hy² – (a – b) xy = 0
∴ A = h, B = -h, 2H = -(a – b)
Equation of the pair of bisectors of (1) is
H(x² – y²) = (A – B) xy
–\(\frac{(a-b)}{2}\)(x² – y²) = 2hxy
-(a – b) (x² – y²) = 4hxy
or (a – b) (x² – y²) + 4hxy = 0
∴ Equation of the pair of bisectors of the pair of bisectors of ax² + 2hyx + by² = 0 is
(a – b) (x² – y²) + 4hxy = 0.

Question 4.
If one line of the pair of lines ax² + 2hxy + by² = 0 bisects the angle between the co-ordinate axes, prove that (a + b)²= 4h².
Solution:
The angular bisectors of the co-ordinate axes are:
y = ±x
Case (i) :
y = x is one of the lines of
ax² + 2hxy + by² = 0
x²(a + 2h + b) = 0
a + 2h + b = 0 ………….. (1)

Case (ii) :
y = – x is one of the lines of
ax² + 2hxy + by² = 0
x² (a – 2h + b) = 0
a – 2h + b = 0 ………. (2)
Multiplying (1) and (2), we get
(a + b + 2h).(a + b – 2h) = 0
(a + b)² – 4h² = 0
(a + b)² = 4h².

Question 5.
If (α, β) is the centroid of the triangle formed by the lines ax² + 2hxy + by² =0 and lx + my = 1, prove that

Solution:
Combined equation of OA, OB is
ax² + 2hxy + by² = 0 …………. (1)
Equation of AB is lx + my = 1
my = 1 – lx
\(y=\frac{1-1 x}{m}\) …………. (2)
Substituting in (1)
ax² + 2hx\(\frac{(1-b)}{m}\) + b\(\frac{(1-1 x)^{2}}{m^{2}}\) = 0
am²x² + 2hmx(1 – lx) + b(1 + l²x² – 2lx) = 0
am²x² + 2hmx – 2hlmx² + b + bl²x² – 2blx = 0
(am² – 2hlm + bl²)x² – 2(bl – hm)x + b = 0
Suppose coordinates of A are (x₁, y₁) and B are (x₂, y₂)
x₁ + x₂ = \(\frac{2(b /-h m)}{a m^{2}-2 h / m+b l^{2}}\) …………. (3)
A and B are points on
lx + my = 1
lx₁ + my₁ = 1
lx₂ + my₂ = 1
l(x₁ + x₂) + m(y₁ + y₂) = 2
m(y₁ + y₂) = 2 – l(x₁ + x₂)

Co-ordinates of the vertices are
O(0, 0), A(x₁, y₁), B(x₂, y₂)
Co-ordinates of G are

Question 6.
Prove that the distance from the origin to the orthocentre of the triangle formed by the lines \(\frac{x}{\alpha}+\frac{y}{\beta}\) = 1 and ax² + 2hxy + by² = 0 is (α² + β²)^1/2 \(\left|\frac{(a+b) \alpha \beta}{a \alpha^{2}-2 h \alpha \beta+b \beta^{2}}\right|\).
Solution:
Let ax² + 2hxy + by² = 0 represent the lines
l₁x + m₁y = 0 ………… (1)
l₂x + m₂y = 0 ………… (2)
∴ (l₁x + m₁y) (l₂x + m₂y) = ax² + 2hxy + by²
Comparing both sides
l₁l₂ = a, m₁m₂ = b, l₁m₂ + l₂ m₁ = 2h
Given line is lx + my =1 ……….. (3)
Clearly the origin O is the point of intersection of (1) & (2)
Let A be the point of intersection of (1) & (3)

By the method of cross multiplication,


Let B be the point of intersection of (2) & (3)
Let P be the orthocentre of ∆ OAB.

⇒ yl₂(l₁α – m₁β) – αβl₁l₂ = m₂(x(l₁α – m₁β) + m₁αβ)
⇒ (l₁α – m₁P) (m₂x – l₂y) = m₁m₂αβ + l₁l₂αβ

Question 7.
The straight line lx + my + n = 0 bisects an angle between the pair of lines of which one is px + qy + r = 0. Show that the other line is (px + qy + r) (l² + m²) – 2(lp + mq) (lx + my + n) = 0.
Solution:
lx + my + n – 0 is a bisector and let (a, P) be any point on it so that
lα + mβ + n = 0 …………….. (1)
The other line will pass through the intersection of given lines and given bisector and hence by p +λq = 0
Its equation is
(px + qy + r) + λ(lx + my + n) = 0 …………….. (2)
Also px + qy + r = 0
If (α, β) be a point on the bisector then its perpendicular distance from the lines (2) and (3) is same.

Putting lα + mβ + n = 0 by (1) in the above and cancelling pα + qβ + r and then squaring both sides, we get
(p + lλ)² + (q + mλ)² = p² + q² or
2λ(pl + qm)+ λ²(l² + m²) = 0
∴ λ = -2\(\frac{p l+Q m}{l^{2}+m^{2}}\)
Substitute X value in (2),
(px + qy + r) + \(\left(\frac{-2(p /+Q m)}{l^{2}+m^{2}}\right)\) lx + my + n = 0
⇒ (px + qy + r)(l² + m²) -2(pl + qm) (lx + my + n ) = 0

Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 1st Year Economics Study Material Textbook Solutions Guide PDF Free Download, TS AP Inter 1st Year Economics Blue Print Weightage 2022-2023, Telugu Academy Intermediate 1st Year Economics Textbook Pdf Download, Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Inter 1st Year Study Material Pdf.

Students can also read AP Inter 1st Year Economics Syllabus & AP Inter 1st Year Economics Important Questions for exam preparation. Students can also go through AP Inter 1st Year Economics Notes to understand and remember the concepts easily.

AP Intermediate 1st Year Economics Study Material Pdf Download | Jr Inter 1st Year Economics Textbook Solutions

AP Inter 1st Year Economics Study Material in English Medium

• Chapter 1 Introduction
• Chapter 2 Theory of Consumers Behaviour
• Chapter 3 Theory of Demand
• Chapter 4 Theory of Production
• Chapter 5 Theory of Value
• Chapter 6 Theory of Distribution
• Chapter 7 National Income
• Chapter 8 Macro Economic Aspects
• Chapter 9 Money, Banking and Inflation
• Chapter 10 Economic Statistics

AP Inter 1st Year Economics Study Material in Telugu Medium

• Chapter 1 పరిచయం
• Chapter 2 వినియోగదారుని ప్రవర్తనా సిద్ధాంతాలు
• Chapter 3 డిమాండ్ సిద్ధాంతాలు
• Chapter 4 ఉత్పత్తి సిద్ధాంతం
• Chapter 5 విలువ సిద్ధాంతం
• Chapter 6 పంపిణీ సిద్ధాంతం
• Chapter 7 జాతీయాదాయం
• Chapter 8 స్థూల ఆర్థిక సిద్ధాంతం
• Chapter 9 ద్రవ్యం, బాంకింగ్, ద్రవ్యోల్బణం
• Chapter 10 ఆర్థిక గణాంక శాస్త్రం

TS AP Inter 1st Year Economics Weightage Blue Print 2022-2023

TS AP Inter 1st Year Economics Weightage 2022-2023 | TS AP Inter 1st Year Economics Blue Print 2022

Intermediate 1st Year Economics Syllabus

TS AP Inter 1st Year Economics Syllabus

Paper – I Introductory Economic Theory

Unit – I Introduction
a) Origin and meaning of Economics –
b) Definitions of Economics; Adam Smith, Alfred Marshall, Lionel Robbins, Paul Samuelson, & Jocob Viner. Concept of Economics – Micro & Macro Economics Deductive and Inductive Method, Static and Dynamic Analysis, Positive and Normative Economics. Goods: (Free, Economic, Consumer, Producer, and Intermediary), Wealth, Income, Utility, Value, Price, wants and welfare.

Unit – II Theory of Consumers Behaviour
a) Cardinal and Ordinal Utility, the law of Diminishing Marginal Utility – Limitations – Importance; law of Equi-Marginal Utility Limitations and –Importance of the Law, Indifference Curve Analysis – Properties and Consumer’s Equilibrium.

Unit – III Theory of Demand
a) Meaning – Demand Function – Determinants of Demand, Demand Schedule – Demand Curve, Law of Demand, Exceptions to Law of Demand – Causes for the downward slope of the demand curve, Types of Demand – Price Demand, Income Demand, and Cross Demand,
b) Elasticity of Demand – Meaning and Types – Price Elasticity, and Income Elasticity and Cross Elasticity – Price Elasticity-Types; Measurement of Price Elasticity of Demand- Point Method. Arc Method, Total Outlay Method. Determinants of Elasticity of Demand; Importance of Elasticity of Demand

Unit – IV Theory of Production
a) Meaning – Production Function – Factors of Production; Short-run and Long-run Production Function; Law of variable proportions
b) Law of returns to scale; Economies of Scale – Internal and External
c) Supply – Supply Function – Determinants of Supply –– Law of Supply
d) Cost Analysis – Basic Concepts of Costs- (Money, Real, Opportunity, Fixed and Variable, Total, Average and Marginal costs)
e) Revenue Analysis – Revenue under perfect and imperfect competition

Unit – V Theory of Value
a) Meaning and Classification of Markets – Perfect competition – features – price determination- Short-run and Long-run equilibrium of a firm and Industry b) Imperfect Competition – Monopoly – Price Determination – Price-Discrimination-Monopolistic Competition- Features- Meaning of Oligopoly – Duopoly

Unit – VI Theory of Distribution
a) Determination of Factor Prices – Marginal Productivity Theory
b) Rent – Ricardian theory of Rent – Modern theory – Quasi Rent – Transfer earnings.
c) Wages – Meaning and types of wages – Money and Real wages
d) Interest- Meaning – Gross and Net interests
e) Profits – Meaning – Gross and Net profits

Unit – VII National Income
a) Definitions of National Income and Concepts
b) Measurement of National Income – Census of Product Method – Census of Income Method – Census of Expenditure Method
c) Methods of Measuring National Income in India; Problems and importance

Unit – VIII Macro Economics Aspects
a) Classical theory of Employment –J.B. Say Law of Markets- Limitations – J.M. Keynes Effective Demand.
b) Public Economics – Public Revenue – Public Expenditure – Public debt – Components of Budget.

Unit – IX Money, Banking and Inflation
a) Money – Definitions and Functions of money – Types of Money
b) Banking – Commercial Banks – Functions; Central Bank – Functions – Reserve Bank of India – Net Banking.
c) Inflation – Definitions – Types – Causes and Effects of inflation – Remedial Measures

Unit – X Statistics for Economics
a) Meaning, Scope and Importance of Statistics in Economics with Diagrams (Bar diagrams and Pie diagrams) Measures of central tendency – Mean, Median, Mode.

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Here is AP Intermediate 1st Year Study Material to help students prepare and ace the AP Intermediate 1st year exams.

AP Inter 1st Year Study Material Pdf

• Intermediate 1st Year Maths 1A Textbook Solutions
• Intermediate 1st Year Maths 1B Textbook Solutions
• AP Inter 1st Year Physics Study Material
• AP Inter 1st Year Chemistry Study Material
• AP Inter 1st Year Botany Study Material
• AP Inter 1st Year Zoology Study Material
• AP Inter 1st Year Telugu Study Material
• AP Inter 1st Year English Study Material
• AP Inter 1st Year Sanskrit Study Material
• AP Inter 1st Year Hindi Study Material
• AP Inter 1st Year Civics Study Material
• AP Inter 1st Year Commerce Study Material
• AP Inter 1st Year Accountancy Study Material
• AP Inter 1st Year Economics Study Material
• AP Inter 1st Year History Study Material

AP Inter 1st Year Important Questions

• Inter 1st Year Maths 1A Important Questions
• Inter 1st Year Maths 1B Important Questions
• AP Inter 1st Year Physics Important Questions
• AP Inter 1st Year Chemistry Important Questions
• AP Inter 1st Year Botany Important Questions
• AP Inter 1st Year Zoology Important Questions
• AP Inter 1st Year Civics Important Questions
• AP Inter 1st Year Commerce Important Questions
• AP Inter 1st Year Accountancy Important Questions
• AP Inter 1st Year Economics Important Questions

AP Inter 1st Year Notes

• Inter 1st Year Maths 1A Formulas
• Inter 1st Year Maths 1B Formulas
• AP Inter 1st Year Physics Notes
• AP Inter 1st Year Chemistry Notes
• AP Inter 1st Year Botany Notes
• AP Inter 1st Year Zoology Notes
• AP Inter 1st Year Civics Notes
• AP Inter 1st Year Commerce Notes
• AP Inter 1st Year Economics Notes
• AP Inter 1st Year History Notes

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Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 1st Year Accountancy Study Material Textbook Solutions Guide PDF Free Download, TS AP Inter 1st Year Accountancy Blue Print Weightage 2022-2023, Telugu Academy Intermediate 1st Year Accountancy Textbook Pdf Download, Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Inter 1st Year Study Material Pdf.

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AP Intermediate 1st Year Accountancy Study Material Pdf Download | Jr Inter 1st Year Accountancy Textbook Solutions

AP Inter 1st Year Accountancy Study Material in English Medium

• Chapter 1 Book Keeping and Accounting
• Chapter 2 Accounting Principles
• Chapter 3 Double Entry Book Keeping System
• Chapter 4 Journal
• Chapter 5 Ledger
• Chapter 6 Subsidiary Books
• Chapter 7 Journal Proper
• Chapter 8 Cash Book
• Chapter 9 Bank Reconciliation Statement
• Chapter 10 Trail Balance
• Chapter 11 Rectification of Errors
• Chapter 12 Final Accounts
• Chapter 13 Final Accounts with Adjustments

AP Inter 1st Year Accountancy Study Material in Telugu Medium

• Chapter 1 పుస్తక నిర్వహణ – గణకశాస్త్రం
• Chapter 2 అకౌంటింగ్ సూత్రాలు
• Chapter 3 జంటపద్దు పుస్తక నిర్వహణ విధానం
• Chapter 4 చిట్టా
• Chapter 5 ఆవర్జా
• Chapter 6 సహాయక చిట్టాలు
• Chapter 7 అసలు చిట్టా
• Chapter 8 నగదు పుస్తకం
• Chapter 9 బ్యాంకు నిల్వల సమన్వయ పట్టిక
• Chapter 10 అంకణా
• Chapter 11 తప్పుల సవరణ
• Chapter 12 ముగింపు లెక్కలు
• Chapter 13 ముగింపు లెక్కలు సర్దుబాట్లు సర్దుబాట్లు

TS AP Inter 1st Year Accountancy Weightage Blue Print 2022-2023

TS AP Inter 1st Year Accountancy Weightage 2022-2023 | TS AP Inter 1st Year Accountancy Blue Print 2022

Intermediate 1st Year Accountancy Syllabus

TS AP Inter 1st Year Accountancy Syllabus

Chapter 1 Book Keeping and Accounting

• 1.1 Introduction
• 1.2 Book Keeping
• 1.3 Accounting
• 1.4 Basic Accounting Terms

Chapter 2 Accounting Principles

• 2.1 Accounting Principles
• 2.2 Accounting Concepts
• 2.3 Accounting Conventions
• 2.4 Accounting Standards

Chapter 3 Double Entry Book Keeping System

• 3.1 Introduction
• 3.2 Meaning
• 3.3 Advantages
• 3.4 Account
• 3.5 Classification of Accounts

Chapter 4 Journal

• 4.1 Meaning
• 4.2 Proforma
• 4.3 Illustrations

Chapter 5 Ledger

• 5.1 Meaning
• 5.2 Advantages of Ledger
• 5.3 Posting

Chapter 6 Subsidiary Books

• 6.1 Meaning of the Subsidiary Books
• 6.2 The Need / Advantages of Subsidiary Books
• 6.3 Types of Subsidiary Books
• 6.4 Preparation of Subsidiary Books
  • 6.4.1 Purchase Book
  • 6.4.2 Purchase Returns Book
  • 6.4.3 Sales Book
  • 6.4.4 Sales Return Book
  • 6.4.5 Cash Book
  • 6.4.6 Bills Receivable Book
  • 6.4.7 Bills Payable Book
  • 6.4.8 Journal Proper

Chapter 7 Journal Proper

• 7.1 Meaning
• 7.2 Advantages
  • 7.2.1 Opening Entries
  • 7.2.2 Purchases of Assets on Credit
  • 7.2.3 Sale of Asset on Credit
  • 7.2.4 Rectification entries
  • 7.2.5 Adjustment entries
  • 7.2.6 Closing journal entries
  • 7.2.7 Transfer entries
  • 7.2.8 Other entries

Chapter 8 Cash Book

• 8.1 Meaning of Cash Book
• 8.2 Characteristics and Advantages
  • 8.2.1 Characteristics
  • 8.2.2 Advantages
• 8.3 Importance
• 8.4 Various kinds of Cash Book and their Preparation
  • 8.4.1 Simple Cash Book
  • 8.4.2 Double column Cash Book
  • 8.4.3 Triple column Cash Book
  • 8.4.4 Petty Cash Book

Chapter 9 Bank Reconciliation Statement

• 9.1 Introduction
• 9.2 Nature Of The Cash Book And Bank Pass Book (Bank Statement)
• 9.3 Meaning and Advantages of Bank Reconciliation Statement
• 9.4 Procedure for Preparation of Bank Reconciliation Statement
• 9.5 Reasons for Difference
• 9.6 Preparation of Bank Reconciliation Statement
  • 9.6.1 Favourable Balances
  • 9.6.2 Unfavourable Balance or Overdraft Balance
  • 9.6.3 When extracts from Cash Book and Pass Book are given

Chapter 10 Trail Balance

• 10.1 Meaning
• 10.2 Features or Characteristics
• 10.3 Merits
• 10.4 Limitations
• 10.5 Types of Preparation
• 10.6 Proforma
• 10.7 Key Points

Chapter 11 Errors and their Rectifications

• 11.1 Errors
• 11.2 Types of Errors
• 11.3 Rectification of Errors
• 11.4 Suspense Account

Chapter 12 Final Accounts

• 12.1 Meaning
• 12.2 Objectives
• 12.3 Advantages and Limitations
• 12.4 Capital and Revenue Items
• 12.5 Preparation of Trading Account
• 12.6 Preparation of Profit & Loss Account
• 12.7 Balance Sheet

Chapter 13 Final Accounts with Adjustments

• 13.1 Meaning
• 13.2 Types of Adjustments
• 13.3 Summary of Adjustments
• 13.4 Accounting Treatment to the Adjustments given in Trial Balance

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AP Intermediate 1st Year Physics Important Questions with Answers Chapter Wise 2022

Intermediate 1st Year Physics Important Questions Chapter Wise 2022

• Chapter 1 Physical World Important Questions
• Chapter 2 Units and Measurements Important Questions
• Chapter 3 Motion in a Straight Line Important Questions
• Chapter 4 Motion in a Plane Important Questions
• Chapter 5 Laws of Motion Important Questions
• Chapter 6 Work, Energy and Power Important Questions
• Chapter 7 Systems of Particles and Rotational Motion Important Questions
• Chapter 8 Oscillations Important Questions
• Chapter 9 Gravitation Important Questions
• Chapter 10 Mechanical Properties of Solids Important Questions
• Chapter 11 Mechanical Properties of Fluids Important Questions
• Chapter 12 Thermal Properties of Matter Important Questions
• Chapter 13 Thermodynamics Important Questions
• Chapter 14 Kinetic Theory Important Questions

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Intermediate Maths 1B Important Questions | Maths 1B Important Questions Chapter Wise with Solutions Pdf 2022

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AP Inter 1st Year Maths 1B Important Questions in English Medium

Jr Inter Maths 1B Important Questions with Answers

1. Inter 1st Year Maths 1B Locus Important Questions
2. Inter 1st Year Maths 1B Transformation of Axes Important Questions
3. Inter 1st Year Maths 1B The Straight Line Important Questions
4. Inter 1st Year Maths 1B Pair of Straight Lines Important Questions
5. Inter 1st Year Maths 1B Three Dimensional Coordinates Important Questions
6. Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Important Questions
7. Inter 1st Year Maths 1B The Plane Important Questions
8. Inter 1st Year Maths 1B Limits and Continuity Important Questions
9. Inter 1st Year Maths 1B Differentiation Important Questions
10. Inter 1st Year Maths 1B Applications of Derivatives Important Questions

AP Inter 1st Year Maths 1B Important Questions in Telugu Medium

• Chapter 1 బిందుపథం Important Questions
• Chapter 2 అక్ష పరివర్తనం Important Questions
• Chapter 3 సరళరేఖ Important Questions
• Chapter 4 సరళరేఖాయుగ్మాలు Important Questions
• Chapter 5 త్రిపరిమాణ నిరూపకాలు Important Questions
• Chapter 6 దిక్ కొసైన్లు, దిక్ సంఖ్యలు Important Questions
• Chapter 7 సమతలం Important Questions
• Chapter 8 అవధులు, అవిచ్ఛిన్నత Important Questions
• Chapter 9 అవకలనం Important Questions
• Chapter 10 అవకలజాల అనువర్తనాలు Important Questions

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Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 1st Year Zoology Study Material Textbook Solutions Guide PDF Free Download, TS AP Inter 1st Year Zoology Blue Print Weightage 2022-2023, Telugu Academy Intermediate 1st Year Zoology Textbook Pdf Download, Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Inter 1st Year Study Material Pdf.

Students can also read AP Inter 1st Year Botany Syllabus & AP Inter 1st Year Zoology Important Questions for exam preparation. Students can also go through AP Inter 1st Year Zoology Notes to understand and remember the concepts easily.

AP Intermediate 1st Year Zoology Study Material Pdf Download | Jr Inter 1st Year Zoology Textbook Solutions

AP Inter 1st Year Zoology Study Material in English Medium

• Chapter 1 Diversity of Living World
• Chapter 2 Structural Organisation in Animals
• Chapter 3 Animal Diversity-I: Invertebrate Phyla
• Chapter 4 Animal Diversity-II: Phylum Chordata
• Chapter 5 Locomotion and Reproduction in Protozoa
• Chapter 6 Biology in Human Welfare
• Chapter 7 Type Study of Periplaneta Americana (Cockroach)
• Chapter 8 Ecology and Environment

AP Inter 1st Year Zoology Study Material in Telugu Medium

• Chapter 1 జీవ ప్రపంచ వైవిధ్యం
• Chapter 2 జంతుదేహ నిర్మాణం
• Chapter 3 జంతు వైవిధ్యం-I: అకశేరుక వర్గాలు
• Chapter 4 జంతు వైవిధ్యం-II: కార్డేటాల క్లుప్త చరిత్ర
• Chapter 5 గమనం, ప్రత్యుత్పత్తి
• Chapter 6 మానవ సంక్షేమంలో జీవశాస్త్రం
• Chapter 7 పెరిప్లానెటా అమెరికానా (బొద్దింక)
• Chapter 8 జీవావరణం – పర్యావరణం

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TS AP Inter 1st Year Zoology Weightage 2022-2023 | TS AP Inter 1st Year Zoology Blue Print 2022

TS AP Inter 1st Year Zoology Weightage to Objectives

TS AP Inter 1st Year Zoology Weightage to Form of Questions

TS AP Inter 1st Year Zoology Weightage to Content: Units

Intermediate 1st Year Zoology Syllabus

TS AP Inter 1st Year Zoology Syllabus

Unit I Zoology – Diversity of Living World

• 1.1 What is life?
• 1.2 Nature, Scope & meaning of zoology
• 1.3 Branches of Zoology
• 1.4 Need for classification- Zoos as tools for the study of taxonomy
• 1.5 Basic principles of Classification: Biological system of classification- (Phylogenetic classification only)
• 1.6 Levels of Hierarchy of classification
• 1.7 Nomenclature – Bi & Trinominal
• 1.8 Species concept
• 1.9 Kingdom Animalia
• 1.10 Biodiversity – Meaning and distribution (Genetic diversity, Species diversity, Ecosystem diversity(alpha, beta, and gamma), other attributes of biodiversity, the role of biodiversity, threats to biodiversity, methods of conservation, IUCN Red data books, Conservation of wildlife in India – Legislation, Preservation, Organisations, Threatened species.

Unit II Structural Organization in Animals

• 2.1 Levels of organization, Multicellularity: Diploblastic & Triploblastic conditions.
• 2.2 Asymmetry, Symmetry: Radial symmetry, and Bilateral symmetry (Brief account giving one example for each type from the representative phyla)
• 2.3 Acoelomates, Pseudocoelomates, and Eucoelomates: Schizo & Entero coelomates (Brief account of the formation of coelom)
• 2.4 Tissues: Epithelial, Connective, Muscular, and Nervous tissues. (make it a little more elaborative)

Unit III Animal Diversity – I: Invertebrate Phyla
General Characters – Strictly restrict to 8 salient features only
Classification up to Classes with two or three examples – Brief account only

• 3.1 Porifera
• 3.2 Cnidaria
• 3.3 Ctenophora
• 3.4 Platyhelminthes
• 3.5 Nematoda
• 3.6 Annelida (Include Earthworm as a type study strictly adhering to NCERT textbook)
• 3.7 Arthropoda
• 3.8 Mollusca
• 3.9 Echinodermata
• 3.10 Hemichordata

Unit IV Animal Diversity – II: Phylum: Chordata
General Characters – Strictly restrict to 8 points only Classification up to Classes – Brief account only with two or three examples

• 4.1 Sub phylum: Urochordata
• 4.2 Sub phylum: Cephalochordata
• 4.3 Sub phylum: Vertebrata
• 4.4 Super Class: Agnatha
• 4.4.1 Class Cyclostomata
• 4.5 Super Class: Gnathostomata
• 4.5.1 Super Class: Pisces
• 4.5.2 Class: Chondrichthyes
• 4.5.3 Class: Osteichthyes
• 4.6 Tetrapoda
• 4.6.1 Class: Amphibia (Include Frog as a type study strictly adhering to NCERT textbook)
• 4.6.2 Class: Reptilia
• 4.6.3 Class: Aves
• 4.6.4 Class: Mammalia

Unit V Locomotion & Reproduction in Protozoa

• 5.1 Locomotion: Definition, types of locomotor structures pseudopodia (basic idea of pseudopodia without going into different types), flagella & cilia (Brief account giving two examples each)
• 5.2 Flagellar & Ciliary movement – Effective & Recovery strokes in Euglena, Synchronal & Metachronal movements in Paramecium
• 5.3 Reproduction: Definition, types. Asexual Reproduction: Transverse binary fission in Paramecium & Longitudinal binary fission in Euglena. Multiple fission, Sexual Reproduction.

Unit VI Biology & Human Welfare

• 6.1 Parasitism and parasitic adaptation
• 6.2 Health and disease: introduction (follow NCERT) Life cycle, Pathogenicity, Treatment & Prevention (Brief account only)
  1. Entamoeba histolytica 2. Plasmodium vivax 3. Ascaris lumbricoides 4. Wuchereria bancrofti
• 6.3 A brief account of pathogenicity, treatment & prevention of Typhoid, Pneumonia, Common cold, & Ringworm.
• 6.4 Drugs and Alcohol abuse

Unit VII Type study of Periplaneta Americana

• 7.1 Habitat and habits
• 7.2 External features
• 7.3 Locomotion
• 7.4 Digestive system
• 7.5 Respiratory system
• 7.6 Circulatory system
• 7.7 Excretory system
• 7.8 Nervous system – sense organs, the structure of ommatidium.
• 7.9 Reproductive system

Unit VIII Ecology & Environment

• 8.1 Organisms and Environment: Ecology, population, communities, habitat, niche, biome, and ecosphere (definitions only)
• 8.2 Ecosystem: Elementary aspects only Abiotic factors – Light, Temperature & Water (Biological effects only), Ecological adaptations
• 8.3 Population interactions
• 8.4 Ecosystems: Types, Components, Lake ecosystem
• 8.5 Food chains, Food web, Productivity and Energy flow in Ecosystem, Ecological pyramids – Pyramids of numbers, biomass, and energy.
• 8.6 Nutrition cycling – Carbon, Nitrogen, & Phosphorous cycles (Brief account)
• 8.7 Population attributes Growth, Natality, and Mortality, Age distribution, and Population regulation.
• 8.8 Environmental issues

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Andhra Pradesh BIEAP AP Inter 1st Year Accountancy Study Material 13th Lesson Final Accounts with Adjustments Textbook Questions and Answers.

AP Inter 1st Year Accountancy Study Material 13th Lesson Final Accounts with Adjustments

Essay Type Questions

Question 1.
Describe the various types of adjustments with examples.
Answer:
Types of Adjustments:
1. Adjustments relating to closing stock: Closing stock means, the stock of goods unsold at the end of the accounting year.
Adjustment entry:

(Being the closing stock transfer to the trading account)

Accounting treatment in final accounts:

1) Show on the credit side of trading A/c

2) Show on the assets side of balance sheet.

Note : If closing stock is given in Trial Balance, show it on the Assets side of Balance sheet.

2. Adjustments relating to expenses:
a) Outstanding expenses : Expenses relating to the current accounting year but not yet paid and are to be paid in the next year e.g: Salary for the month of December is due but not paid.

Adjustment entry:

(Being the expenses due)

Accounting treatment in final accounts:

1) Add either in trading A/c or in profit & loss A/c to the concerned expenditure item.
Trading A/c

2) Show it as a liability on the liabilities side of Balance sheet.

Note : If outstanding expenses are given in trial balance show as liability in Balance sheet.

b) Prepaid expenses : Expenses relating to the next accounting year but paid in the current accounting period are called prepaid expenses. (May. ’17 – A.P.)

Adjustment entry:

(Being expenses paid in advance)

Accounting treatment in final accounts: If prepaid expenses are given as an adjustment.

1. Deduct it from the concerned expenditure either in trading A/c or in Profit & Loss A/c for the first instance and
2. Record as asset on assets side of the balance sheet as second time.

1) Add either in trading A/c or in profite & loss A/c to the concerned expenditures item.

Trading A/c

Balance Sheet

Note: If prepaid expenditure is given only in Trial balance, show it as asset in Balance sheet.

3. Income:
a) Accrued Income: Income relating to current year which is not received during the current year but to be received in the next year is called Accured income or income receivable.
Adjustment entry:

(Being the income receivable)

Accounting treatment in final accounts: If accrued income is given as adjustment –

1. For the first instance add to the concerned income in profit and loss a/c on credit side and then.
2. Show it as an asset in balance sheet on assets side.

Profit & Loss A/c

Balance Sheet

Note : If accrued income is given in trial balance, show it on assets side of Balance sheet.

b) Income Received in Advance : The income relating to the next year but received in the current year is called income received in advances.
Adjustment entry:

(Being the income received in advance)

Accounting treatment in final accounts: When income received in advance is given adjustment

1. Deduct it from the concerned income in Profit & Loss a/c on credit side and
2. Record it as a liability on the liabilities side in the balance sheet.

Profit & Loss A/c

Balance Sheet

Note: If Income received in advance is given in the trial balance show it on liabilities side in the balance sheet.

4. Depreciation: Decline in the value of fixed assets is called “Depreciation”.
Adjustment entry:

(Being the depreciation provided on asset)

Accounting treatment in final accounts: When depreciation is given as an adjustment:

1. Debit it to profit & loss A/c.
2. Deduct it from the value of concerned asset in balance sheet on assets side.

Trial Balance

Adj : Provide depreciation on machinery 10%

Profit & Loss A/c

Note : If depreciation is given in trial balance, it should be shown on debit side in P & L A/c only.

5. Debtors : In final accounts bad debts, provision for bad debts may be given as adjustments relating to debtors.
A) Bad debts: To debts which are not collected or irrecoverable are known as bad debts.

Adjustment entry:

(Being bad debts written off)

Accounting treatment in final accounts:

a) When bad debts are given, only in the adjustments –

1. Debit to profit & loss A/c and
2. Deduct from debtors in the balance sheet on assets side.

Trial Balance

Adjustment: Bad debts : 500

Note : If the bad debts are given in trial balance only, it should be shown on debit side in Profit & Loss A/c.

b) When Bad debts are given in both Trial Balance and adjustments:

1. In Profit & Loss A/c, both the bad debts (Bad debts given in Trial balance and given in adjustment) are to be shown on debit side.
2. Bad debts given only in the adjustments are to be deducted from debtors in the balance sheet.

Trial Balance

Adjustments: 1) Bad debts : 400

B) Provision for bad and doubtful debts: Some debts of a particular year may or may not be recovered in the next year. These debts are known as doubtful debts. So traders create same amount on current year debtors and keep the same to meet the doubtful bad debts of the next year, which is called provision for bad and doubtful debts.

a) When provision for doubtful debts is given as adjustment:
Adjustment entry:

(Being provision created on debtors)

Accounting treatment in final accounts:

1. Show it on debit side in profit & Loss A/c and
2. Deduct it from debtors in Balance sheet,

e.g.:
Trial Balance

Adjustment: Create provision for bad and doubtful debts 5%.

Profit & Loss A/c

Balance Sheet

b) When provision for doubtful debts is given in Trial Balance and also in adjustments: Accounting treatment in final accounts:

1. Compare the old provision (given in trial balance) with new provision (given in the adjustments), if the new provision is more than the old provision, the difference amount (New provision – old provision) should be debited to the Profit & Loss A/c.
On the other hand, new provision is less than the old provision, the difference amount (old provision – new provision) should be recorded on the credit side of Profit & Loss A/c.

2. In balance sheet, deduct the amount of new provision of bad and doubtful debts from sundry debtors.

Trial Balance

Adjustments : Create 5% provision for doubtful debts.

Profit & Loss A/c

Balance Sheet

c) If the bad debts are given both in trial balance and in adjustments, and also provision for bad debts given in adjustments.
Accounting treatment in final accounts:

1. Don’t calculate the provision directly on sundry debtors.
2. Calculate the provision after deducting the further bad debts.

Trial Balance

Adjustments:

1. Further bad debts : Rs. 600
2. Provision for bad debts : 5%

Profit & Loss A/c

6. Interest on capital : It is the amount of interest payable on owner’s capital by the business organisation.

Adjustment entry:

(Being the interest payable on capital)

Accounting treatment in final A/cs:

1. Debit in profit & Loss A/c and
2. It should be added to the capital in balance sheet.

Trial Balance

Adjustment: Interest on capital: 12%

7. Interest on Drawings : Drawings mean the amount of cash or goods taken by the trader for personal use. The amount of interest payable by the owner to the business is called interest on drawings.
Adjustment entry:

(Being the interest on drawings)
Accounting treatment in final A/cs:

1. It is to be recorded on credit side of P & L a/c and
2. It should be deducted from capital in balance sheet.

Trial Balance

Adjustment: Interest on drawings : 5%

Note: When interest on drawings is given in trial balance, it should be shown on credit side in Profit & Loss A/c only.

Short Answer Questions

Question 1.
Write the following:
a) Interest on Capital:
Answer:
The amount of interest payable on owner’s capital by the business organisation is called interest on capital.
Adjustment entry:

(Being the interest payable on capital)
Accounting treatment in final accounts:
When interest on capital is given as an adjustment.
1. Debit in P & L A/c and
2. It should be added to the capital in balance sheet.
Note : When it is given in trial balance, debit it in P & L A/c only.

b) Interest on Drawings :
Answer:
Drawings mean the amount of cash or goods taken by the trader for personal use.
The amount of interest payable by the owner to the business is called Interest on drawings.
Adjustment entry:

(Being the interest on drawings)
Accounting treatment In final accounts:

When interest on drawings given as adjustment.

1. It is to be recorded on credit side of P & L A/c and
2. Deduct the amount from capital in Balance sheet.

Note: When interest on drawings is given in trial balance, it should be shown on credit side in Profit & Loss A/c.

Very Short Answer Questions

Question 1.
What is the meaning of adjustment ?
Answer:
To find out net profit and true financial position, all expenses relating to current year whether actually paid or not, all incomes received or yet to be received should be taken into account. Some of the incomes and expenses relating to next year, but received and paid in the current year should not be included in the accounts of current year. The amount to be adjusted to the concerned items is called adjustment. e.g: Outstanding salaries, prepaid insurance, etc.

Question 2.
Explain the importance of adjustment:
Answer:

1. Expenses or incomes relating to the accounting period can be known accurately.
2. Profit or loss can be ascertained accurately.
3. Real value of assets and liabilities can be ascertained easily.

Question 3.
Give the meaning of bad debts. (Mar. 2018 T.S.)
Answer:
The debts which are not collected or Irrecoverable are known as bad debts.
Adjuštment entry:

(Being bad debts written off)

Adjustments Summary

Problems

Question 1.
From the following trial balance, prepare final accounts of Praveen Traders as on 31.12.2013:

Adjustments:

1. Closing stock: 4500;
2. Outstanding wages : 390;
3. Outstanding salaries : 500
4. Prepaid Insurance: 400

Answer:

Question 2.
From the following particulars, prepare final accounts : (May ’17 – T.S.)

Adjustments:

1. Closing stock: 6000
2. Prepaid Insurance: 200
3. Outstanding salaries :600
4. Accrued interest : 500

Answer:

Question 3.
From the following particulars, prepare final accounts of Giri for the year ending 31.12.2013.

Adjustments:

1. Closing stock value: 3500
2. Outstanding wages : 860
3. Prepaid insurance: 100
4. Provide depreciation on furniture: 10% and on land & buildings : 10%
5. Interest received in advance : 500

Answer:

Question 4.
From the following Trialbalance o1 Mr.kapil, prepare Trading P & L A/c and Balance Sheet or the year ended (Mar. 2018 – A.P.)
Trial Balance

Adjustments:

1. Outstanding wages: 2000;
2. Outstanding salaries: 1000;
3. Prepaid insurance: 50;
4. Create 5% reserve for bad debts on debtors;
5. Depreciation on furniture: 150, Dep. on machinery: 500;
6. Closing stock: 11,000.

Answer:

Question 5.
From the following particulars, prepare final accounts for the year ended 31.3.2010.
Trial Balance

Adjustments:

1. Closing stock: 16,800;
2. Interest on capital :9%;
3. Write off : 2,000 as bad debt and provide 5% reserve for doubtful debts;
4. Outstanding wages: 1,000.

Answer:

Question 6.
Prepare final accounts of Praveen Traders for the year ending 31.03.2014.
Trial Balance

Adjustments:

1. Closing stock : 5,800;
2. Depreciation on motor van: 10%;
3. Reserve for bad & doubtful debts : 5%;
4. Outstanding rent Rs. 500;
5. Prepaid taxes: Rs. 200/-.

Answer:

Question 7.
Prepare final accounts from the following trial balance for the year ended 31.12.2013.
Trial Balance

Adjustments:

1. Closing stock: 2,100
2. Outstanding stationery bill : 600
3. Depreciation on machinery: 10%
4. Bad Debts : 500
5. Prepaid wages :500

Answer:

Question 8.
From the following Trial balance of Vinod Traders, prepare final accounts:
Trial Balance

Adjustments:

1. Closing stock: 9,500
2. Bad debts : 1,500
3. Provide reserve for bad debts : 5%
4. Outstanding wages : 300
5. Depreciation on machinery: 10%
6. Interest received in advance : 500.

Answer:

Question 9.
Prepare sole traders final accounts for the year ending 31.03.2014.
Trial Balance

Adjustments:

1. Closing stock value : 7,500;
2. Depreciation on machinery : 12%;
3. Commission received in advance : 1,200;
4. Interest receivable : 1,500;
5. Further bad debts : 400;
6. Prepaid insurance: 500.

Answer:

Question 10.
Prepare Final Accounts of Ramakrishna Traders as on 31.12.2013:

Adjustments:

1. Closing stock: 3,500
2. Outstanding rent: 500
3. Prepaid salaries & wages : 400
4. Interest received in advance: 300
5. Depreciation on machinery: 10%

Answer:

Question 11.
Prepare Ravi Traders’ Final Accounts for the fear ended 31.12.2013:
Trial Balance

Adjustments:

1. Closing Stock Value : 5,100
2. Reserve for Bad Debts : 5%
3. Depreciation on patents : 20%
4. Outstanding Rent :300
5. Commission Receivable : 200

Answer:

Question 12.
Prepare Final Accounts of Srinivasa Traders as on 31.12.2012.
Trial Balance

Adjustments:

1. Closing stock value: Rs. 5,000
2. Calculate Interest on Capital : 8%
3. Interest on Drawings: 10%
4. Provide Reserve for Debts : 5%
5. Depredation on premises: 10%

Answer:

Question 13.
From the following Trial Balance prepare Final Accounts.
Trial Balance

Adjustments:

1. Closing Stock Value : Rs. 16,800;
2. Outstanding Salaries : 400
3. Prepaid Rent & Taxes: 201
4. Provide Reserve on Sundry Debtors : 5%
5. Depreciation on Machinery: 10%
6. Interest on Capital: 5%

Answer:

Question 14.
From the following Trial Balance of Vishnu traders prepare Final Accounts for the year ended 31.3.2014.
Trial Balance

Adjustments

1. Closing Stock Value: Rs. 14,000;
2. Depreciation on Furniture: 250, on Machinery: 750
3. Outstanding Wages : Rs. 500;
4. Bad Debts : 600;
5. Interest on Drawings : 5%

Answer:

Question 15.
Prepare Final Accounts:
Trial Balance

Adjustments:

1. Closing Stock Value : Rs. 56,000
2. Outstanding Salaries : 6,000
3. Bad Debts : 2000, and Create Reserve for Bad debts : 3%
4. Depreciation on Machinery: 5%
5. Interest on Capital: 5%

Answer:

Question 16.
From the following Trial Balance and additional information of Latha, prepare Trading and Profit and Loss Account for the year ended 3l Dec. 2008 and Balance Sheet as on that date.
Trial Balance

Adjustments:

1. Closing Stock : Rs. 26,800
2. Depreciate 10% on Machinery and 20% on Patents
3. Outstanding Salaries : Rs. 1,500
4. Unexpired Insurance: Rs. 170
5. Provide 5% provision for bad debts on Debtors

Answer:

Question 17.
From the following Trial Balance of Mr. Paramesh, prepare the Trading, Profit and Loss account and Balance Sheet for the year ended 31.12.2012.
Trial Balance as on 31.12.2012

Adjustments:

1. Closing Stock : Rs. 34,500
2. Outstanding salaries : Rs. 5,500
3. Depreciate plant and machinery by 5%
4. Prepaid insurance: Rs. 1,500
5. 5% provision is to be made for bad debts on debtors

Answer:

Student Activity

Visit any organisation and note the adjustments made during the last year’s final accounts.

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Inter 1st Year Telugu Poems పద్య భాగం

• Poem 1 ధర్మ పరీక్ష
• Poem 2 తిన్నని ముగ్ధ భక్తి
• Poem 3 నందీశ్వరుని శాపం
• Poem 4 శ్మశానవాటి
• Poem 5 చేయెత్తి జైకొట్టు తెలుగోడా!
• Poem 6 కపాలమోక్షం

Inter 1st Year Telugu Textbook Lessons గద్య భాగం

• Chapter 1 హాసము – హాస్యము
• Chapter 2 అప్పుడు పుట్టి ఉంటే…
• Chapter 3 మహిళోద్యమ జనకుడు : కందుకూరి
• Chapter 4 కలవారి కోడలూ కలికి కామాక్షి
• Chapter 5 గాన సుధాకరుడు – మంగళంపల్లి
• Chapter 6 మాటతీరు

Inter 1st Year Telugu Non-Detailed ఉపవాచకం కథా నందనం

• Chapter 1 కుంకుడాకు
• Chapter 2 అంపకం
• Chapter 3 ఊతకర్ర
• Chapter 4 సౌందర్యం
• Chapter 5 ఎచ్చరిక
• Chapter 6 దహేజ్

Inter 1st Year Telugu Grammar

Intermediate 1st Year Telugu Grammar Pdf

• సంధులు
• సమాసాలు
• లేఖారచన
• పదదోషాలు
• స్థూల అవగాహన
• అనువాదం

Intermediate 1st Year Telugu Syllabus

AP Inter 1st Year Telugu Syllabus

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