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

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

I.

Question 1.

Find the angle between the lines represented by 2x² + xy – 6y² + 7y – 2 = 0.

Solution:

Comparing with

ax² + 2hxy + by² + 2gx + 2fy + c = 0

a = 2, 2g = 0, g = 0

b = – 6, 2f = 7, f = 7/2

c = – 2, 2h = 1, h = 1/2

Angle between the lines is given by

Question 2.

Prove that the equation 2x² + 3xy – 2y² + 3x + y + 1 = 0 represents a pair of perpendicular lines.

Solution:

Given a = 2, b = -2

a + b = 0 ⇒ cos α = 0 ⇒ α = π/2

∴ The given lines are perpendicular.

II.

Question 1.

Prove that the equation 3x² + 7xy + 2y² + 5x + 5y + 2 = 0 represents a pair of straight lines and find the co-ordinates of the’point of intersection.

Solution:

The given equation is

3x² + 7xy + 2y² + 5x + 5y + 2 = 0

Comparing a = 3 2f = 5 ⇒ f = \(\frac{5}{2}\)

b = 2 2g = 5 ⇒ g = \(\frac{5}{2}\)

c = 2 2h = 7 ⇒ h = \(\frac{7}{2}\)

∆ = abc + 2fgh – af² – bg² – ch²

= 3(2)(2) + 2.\(\frac{5}{2}\).\(\frac{5}{2}\).\(\frac{7}{2}\) – 3.\(\frac{25}{4}\) – 2.\(\frac{25}{4}\) – 2.\(\frac{49}{4}\) = 0

= \(\frac{1}{4}\)(48 + 175 – 75 – 50 – 98)

= \(\frac{1}{2}\)(223 – 223) = 0

∴ The given equation represents a pair of lines point of intersection is

Question 2.

Find the value of k, if the equation 2x² + kxy – 6y² + 3x + y + l = 0 represents a pair of straight lines. Find the point of intersection of the lines and the angle between the straight lines for this value of k.

Solution:

The given equation is

2x² + kxy – by² + 3x + y+ 1 = 0

a = 2 2f = 1 ⇒ f = \(\frac{1}{2}\)

b = -6 2g = 3 ⇒ g = \(\frac{3}{2}\)

c = 1 2h = k ⇒ h = \(\frac{k}{2}\)

The given equation represents a pair of straight lines abc + 2fgh – af² – bg² – ch² = 0

-12 + 2.\(\frac{1}{2}\).\(\frac{3}{2}\)(+\(\frac{k}{2}\) -2.\(\frac{1}{4}\) + 6.\(\frac{9}{4}\) – \(\frac{k^{2}}{4}\) = 0

– 48 + 3k – 2 + 54 – k² = 0

-k² + 3k + 4 = 0 ⇒ k² – 3k – 4 = 0

(k – 4) (k + 1) = 0

k = 4 or – 1.

Case (i): k = – 1

Point of’intersection is \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

Point of intersection is (\(\frac{-5}{7}\), \(\frac{1}{7}\))

Angle between the lines

Case (ii): k = 4

Point of intersection is P(-\(\frac{5}{8}\), –\(\frac{1}{8}\))

Question 3.

Show that the equation x² – y² – x + 3y – 2 = 0 represents a pair of perpendicular lines and find their equations.

Solution:

Comparing a= 1, f = \(\frac{3}{2}\)

b = -1, g = –\(\frac{1}{2}\)

c = -2, h = 0

abc + 2fgh – af² – bg² – ch²

= 1 (-1) (-2) + 0- 1.\(\frac{9}{4}\) + 1.\(\frac{1}{4}\) + 0

= + 2 – \(\frac{9}{4}\) + \(\frac{1}{4}\) = 0

h² – ab = 0 – 1(-1) = 1 > 0,

f² – be = \(\frac{9}{4}\) – 2 = \(\frac{1}{4}\) = 1 > 0

g² – ac = \(\frac{1}{4}\) + 2 = \(\frac{9}{4}\) > 0

a + b = 1 – 1 = 0

The given equation represent a pair of per-pendicular lines

Let x² – y² – x + 3y – 2

= (x + y + c_{1}) (x – y + c_{2})

Equating the co-efficients of x

⇒ c_{1} + c_{2} = – 1

Equating the co-efficients of y

⇒ – c_{1} + c_{2} = 3

Adding 2c_{2} = 2 ⇒ c_{2} = 1

c_{1} + c_{2} = – 1 ⇒ c_{1} + 1 = – 1

c_{1} = – 2

Equations of the lines are x + y – 2 = 0 and x – y + 1 = 0

Question 4.

Show that the lines x² + 2xy – 35y² – 4x + 44y – 12 = 0 are 5x + 2y – 8 = 0 are concurrent.

Solution:

Equations of the given lines are

x² + 2xy- 35y² -4x + 44y- 12 = 0

a = 1, f = 22

b = – 35, g = – 2

c = – 12, h = 1

Point or intersection is \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

Point of intersection of the given lines is P(\(\frac{4}{3}\), \(\frac{2}{3}\))

5x + 2y – 8 = 5.\(\frac{4}{3}\) + 2.\(\frac{2}{3}\) – 8

= \(\frac{20+4-24}{3}\) = 0

P lies on the third line 5x + 2y – 8 = 0

∴ The given lines are concurrent.

Question 5.

Find the distances between the following pairs of parallel straight lines :

i) 9x² – 6xy + y² + 18x – 6y + 8 = 0

Solution:

Distance between parallel lines = \(2 \sqrt{\frac{g^{2}-a c}{a(a+b)}}\)

ii) x² + 2√3xy + 3y² – 3x – 3√3y – 4 = 0

Solution:

Distance between parallel lines = \(2 \sqrt{\frac{g^{2}-a c}{a(a+b)}}\)

Question 5.

Show that the two pairs of lines 3x² + 8xy – 3y² = 0 and 3x² + 8xy – 3y² + 2x – 4y – 1 = 0 form a square.

Solution:

Combined equation of CA and CB is

3x² + 8xy – 3y²

(x + 3y) (3x – y) = 0

3x – y = 0, x + 3y = 0

Equation of OA is 3x – y = 0 ………. (1)

Equation of OB is x + 3y = 0 ……….(2)

Combined equation of CA and CB is

3x² + 8xy – 3y² + 2x – 4y + 1 =0

Let 3x² + 8xy – 3y² + 2x – 4y + 1 = (3x – y + c_{1}) (x + 3y + c_{2})

Equating the co-efficients of

x, we get c_{1} + 3c_{2} = 2

y, we have 3c_{1} + c_{2} = -4

Equation of BC is 3x – y – 1 = 0 ………. (3)

Equation of AC is x + 3y + 1 = 0 ………. (4)

OA and BC differ by a constant ⇒ OA parallel to BC

OB and CA differ by a constant ⇒ OB parallel to AC

From combined equation of OA and OB

OACB is a rectangle a + b = 3 – 3 = 0

OA = Length of the -L lar from O to AC = \(\frac{|0+0+1|}{\sqrt{1+9}}=\frac{1}{\sqrt{10}}\)

OB = Length of the perpendicular from O to BC = \(\frac{|0+0-1|}{\sqrt{9+1}}=\frac{1}{\sqrt{10}}\)

OA = OB and OACB is a rectangle

OACB is a square.

III.

Question 1.

Find the product of the lengths of the perpendiculars drawn from (2, 1) upon the lines

12x² + 25xy + 12y² + 10x + 11y + 2 = 0

Solution:

Combined equation of AB, AC is

12x² + 25xy+ 12y² + 10x + 11y + 2 = 0

12x² + 25xy +12y²

= 12x² + 16xy + 9xy + 12y² = 0

= 4x (3x + 4y) + 3y (3x + 4y)

= (3x + 4y) (4x + 3y)

Let 12x² + 25xy + 12y² + 10x + 11y + 2

= (3x + 4y + c_{1}) (4x + 3y + c_{2})

Equating the co-efficients of x, we get

4c_{1} + 3c_{2} = 10 ………. (1)

Equating the co-efficients of y, we get

3c_{1} + 4c_{2} = 11 ………… (2)

i.e., 4c_{1} + 3c_{2} – 10 = 0

3c_{1} + 4c_{2} – 11 = 0

Equation of AB is 3x + 4y + 1 = 0

Equation of AC is 4x + 3y + 2 = 0

PQ = Length of the perpendicular from P(2, 1) on

AB = \(\frac{6+4+1}{\sqrt{9+16}}=\frac{11}{5}\)

PQ = Length of the perpendiculars from P(2, 1) on

AC = \(\frac{|8+3+2|}{\sqrt{16+9}}=\frac{13}{5}\)

Product of the length of the perpendiculars

= PQ × PR= \(\frac{11}{5}\) × \(\frac{13}{5}\) = \(\frac{143}{25}\)

Question 2.

Show that the straight lines y² – 4y + 3 = 0 and

x² + 4xy + 4y² + 5x + 10y + 4 = 0 form a parallelogram and find the lengths of its sides.

Solution:

Equation of the first pair of lines is

y² – 4y + 3 = 0

(y – 1) (y – 3) = 0

y – 1 = 0 or y – 3 = 0

Equation of AB is y – 1 = 0 ……….. (1)

Equation of CD is y – 3 = 0 ………. (2)

Equations of AB and CD differ by a constant.

∴ AB and CD are parallel.

Equation of the second pair of lines is

x² + 4xy + 4y² + 5x + 10y + 4 = 0

(x + 2y)² + 5(x + 2y) + 4 = 0

(x + 2y)² + 4 (x + 2y) + (x + 2y) + 4 = 0

(x + 2y)(x + 2y + 4) + 1(x + 2y + 4) = 0

(x + 2y + 1) (x + 2y + 4) = 0

x + 2y + 1 = 0, x + 2y + 4 = 0

Equation of AD is x + 2y + 1 = 0 ……… (3)

Equation of BC is x + 2y + 4 = 0 ……… (4)

AD and BC are parallel.

Solving (1), (3) x + 2 + 1 = 0

x = – 3

Co-ordinates of A are (-3, 1)

Solving (2), (3) x + 6 + 1 = 0

x = -7

Co-ordinates of DC are (-7, 3)

Solving (1), (4) x + 2 + 4 = 0

x = – 6

Co-ordinates of B are (-6, 1)

Lengths of the sides of the parallelogram are 3, 2√5.

Question 3.

Show that the product of the perpendicular distances from the origin to the pair of straight lines represented by ax² + 2hxy + by² + 2gx + 2fy + c = 0 is \(\frac{|c|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

Solution:

Let ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents the lines

l_{1}x + m_{1}y + n_{1} = 0 ……….. (1)

l_{2}x + m_{2}y + n_{2} = 0 ………. (2)

⇒ ax² + 2hxy + by² + 2gx + 2fy + c ≡ (l_{1}x + m_{1}y + n_{1}) (l_{2}x + m_{2}y + n_{2})

l_{1}l_{2} = a, m_{1}m_{2} = b, l_{1}m_{2} + l_{2}m_{1} = 2h,

l_{1}n_{2} + l_{2}n_{1} = 2g, m_{1}n_{2} + m_{2}n_{1} = 2f, n_{1}n_{2} = c

⊥^{Ir} distance from origin to (1) is = \(\frac{\left|n_{1}\right|}{\sqrt{l_{1}^{2}+m_{1}^{2}}}\)

⊥^{Ir}distance from origin to (2) is = \(\frac{\left|n_{2}\right|}{\sqrt{I_{2}^{2}+m_{2}^{2}}}\)

Product of perpendiculars

Question 4.

If the equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of intersecting lines, then show that the square of the distance of their point of intersection from the origin is \(\frac{c(a+b)-f^{2}-g^{2}}{a b-h^{2}}\). Also slow that the square of this distance is \(\frac{f^{2}+g^{2}}{h^{2}+b^{2}}\) if the given lines are perpendicular.

Solution:

Let the equation

ax² + 2hxy + by² + 2gx + 2fy + c = 0

represent the lines

l_{1}x + m_{1}y + n_{1} = 0 …….. (1)

l_{2}x + m_{2}y + n_{2} = 0 …….. (2)

(l_{1}x + m_{1}y + n_{1})(l_{2}x + m_{2}y + n_{2}) = ax² + 2hxy + by² + 2gx + 2fy + c

l_{1}l_{2} = a, m_{1} m_{2} = b, n_{1}n_{2} = c

l_{1}m_{2} + l_{2}m_{1} = 2h, l_{1}n_{2} + l_{12}n_{1} = 2g, m_{1}n_{2} + m_{2}n_{1} = 2f

Solving (1) and (2)

\(\frac{x}{m_{1} n_{2}-m_{2} n_{1}}=\frac{y}{I_{2} n_{1}-l_{1} n_{2}}=\frac{1}{l_{1} m_{2}-I_{2} m_{1}}\)

The point of intersection, P

If the given pair of lines are perpendicular, then a + b = 0

∴ a = -b

\(\mathrm{OP}^{2}=\frac{0-r^{2}-g^{2}}{(-b) b-h^{2}}=\frac{r^{2}+g^{2}}{h^{2}+b^{2}}\)