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 ax^{2} + 2hxy + by^{2} = 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 ax^{2} + 2hxy + by^{2} = 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 h^{2} = ab, then ax^{2} + 2hxy + by^{2} = 0 represents coincident or parallel lines.

→ ax^{2} + 2hxy + by^{2} = 0 represents a pair of ⊥^{lr} lines ⇒ a+ b = 0 i.e., coeft. of x^{2} + coeff. of y^{2} = 0.

→ The equation of pair of lines passing through origin and perpendicular to ax^{2} + 2hxy + by^{2} = 0 is bx^{2} – 2hxy + ay^{2} = 0.

→ The equation of pair of lines passing through (x_{1}, y_{1}) and perpendicular to ax^{2} + 2hxy + by^{2} = 0 isb (x – x_{1})^{2} – 2h(x – x_{1}) (y – y_{1}) + a(y – y_{1})^{2} = 0.

→ The equation of pair of lines passing through (x_{1}, y_{1}) and parallel to ax^{2} + 2hxy + by^{2} = 0 is a(x-x,)^{2} + 2h(x-x1)(y-y1) + b(y – y_{1})^{2} = 0.

→ The equations of bisectors of angles between the lines a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 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 ax^{2} + 2hxy + by^{2} = 0 is h(x^{2} – y^{2}) = (a – b)xy.

→ The product of the perpendicular is from (α, β) to the pair of lines ax^{2} + 2hxy + by^{2} = 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 ax^{2} + 2hxy + by^{2} = 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 ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represents a pair of lines, then

- abc + 2fgh – af
^{2}– bg^{2}– ch^{2}= 0 - h
^{2}≥ ab - g
^{2}≥ ac - f
^{2}≥ be

→ If ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represents a pair of lines and h^{2} > 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 ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represents a pair of parallel lines then h^{2} = ab and af^{2} = bg^{2}

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 L_{1} = 0, L_{2} = 0 be the equations of two straight lines. If P(x_{1}, y_{1}) is a point on L_{1} then it satisfies the equation L_{1} = 0. Similarly, if P(x_{1}, y_{1}) is a point on L_{2} = 0 then it satisfies the equation.

If P(x_{1}, y_{1}) lies on L_{1} or L_{2}, then P(x_{1},y_{1}) satisfies the equation L_{1}L_{2}= 0.

L_{1}L_{2} = 0 represents the pair of straight lines L_{1} = 0 and L_{2} = 0 and the joint equation of L_{1} = 0 and L_{2} = 0 is given by L_{1} L_{2}= 0. ……………(1)

On expanding equation (1) we get and equation of the form ax^{2} + 2hxy + by^{2} + 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 ax^{2} + 2hxy + by^{2} = 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 ax^{2} + 2hxy + by^{2} + 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 h^{2} ≥ ab then ax^{2} + 2hxy + by^{2} = 0 represents a pair of straight lines passing through the origin.

Proof:

Case (i) : Suppose a = 0.

Given equation ax^{2} + 2hxy + by^{2} = 0 reduces to 2hxy + by^{2} = 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 ax^{2} + 2hxy + by^{2} = 0

⇒ a^{2}x^{2} + 2ahxy + aby^{2} = 0

⇒ (ax)^{2} + 2(ax)(hy) + (hy)^{2} – (h^{2} – ab)y^{2} = 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 h
^{2}> ab , the two lines are distinct. - If h
^{2}= ab , the two lines are coincident. - If h
^{2}< ab , the two lines are not real but intersect at a real point (the origin). - If the two lines represented by ax
^{2}+ 2hxy + by^{2}= 0 are taken as l_{1}x + m_{1}y = 0 and l_{2}x + m_{2}y = 0 then

ax^{2}+ 2kxy + by^{2}= (4 x + m_{1}y) (x + m_{2}y) = 2 x^{2}+ (l_{1}m_{2}+ l_{2}m_{1}) xy + m_{1}m_{2}y^{2} - Equating the coefficients of x
^{2}, xy and y^{2}on both sides, we get l_{1}l_{2}= a, l_{1}m_{2}+ l_{2}m_{1}= 2h , m_{1}m_{2}= b.

Theorem:

If ax^{2} + 2hxy + by^{2} = 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 ax^{2} + 2hxy + by^{2} = 0 represent the lines l_{1}x + m_{1}y = 0 ………….(1) and l_{2}x + m_{2}y = 0 ………….(2).

Then l_{1}l_{2} = a, l_{1}m_{2} + l_{2}m_{1} = 2h , m_{1}m_{2} = 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 ax^{2} + 2hxy + by^{2} = 0, then cos θ = ±\(\frac{a+b}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

Proof:

Let ax^{2} + 2hxy + by^{2} = 0 represent the lines l_{1} x + m_{1} y = 0 ………..(1) and l_{2}x + m_{2} y = 0 …………..(2).

Then l_{1}l_{2} = a, l_{1}m_{2} + l_{2}m_{1} = 2h , m_{1}m_{2} = 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 ax^{2} + 2hxy + by^{2} = 0 then cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

If θ is the accute angle between the lines ax^{2} + 2hxy + by^{2} = 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 ax
^{2}+ 2hxy + by^{2}= 0 are perpendicular to each other then θ = π/ 2 and cos θ = 0 ⇒ a + b = 0 i.e., co-efficient of x^{2}+ coefficient of y^{2}= 0. - If the two lines are parallel to each other then 0 = 0.

⇒ The two lines are coincident ⇒ h^{2}= ab

**Bisectors of Angles:**

Theorem:

The equations of bisectors of angles between the lines a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 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 ax^{2} + 2hxy + by^{2} = 0 is h(x^{2} – y^{2}) = (a – b)xy (or) \(\frac{x^{2}-y^{2}}{a-b}=\frac{x y}{h}\).

Proof:

Let ax^{2} + 2hxy + by^{2} = 0 represent the lines l_{1}x + m_{1}y = 0 ……….(1)

and l_{2}x + m_{2}y = 0 ………(2).

Then l_{1}l_{2} = a, l_{1}m_{2} + l_{2}m_{1} = 2h , m_{1}m_{2} = 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 (x_{0}, y_{0}) and parallel ax^{2} + 2hxy + by^{2} = 0

is a( x – x_{0})^{2} + 2h( x – x_{0})(y – y_{0}) + b( y – y_{0})^{2} = 0

Proof:

Let ax^{2} + 2hxy + by^{2} = 0 represent the lines l_{1}x + m_{1}y = 0 ……(1) and l_{2}x + m_{2} y = 0 …….. (2).

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

The equation of line parallel to (1) and passing through (x_{0}, y_{0}) is l_{2}(x – x_{0}) + m_{2}(y – y_{0}) = 0 ………(3)

The equation of line parallel to (2) and passing through (x_{0}, y_{0}) is l_{2}(x – x_{0}) + m_{2}(y – y_{0}) = 0 ………(4)

The combined equation of (3), (4) is

[l_{1}( x – x_{0}) + m_{1}( y – y_{0})][l_{2}( x – x_{0}) + m_{2}(y – y_{0})] = 0

⇒ l_{1}l_{2}(x – x_{0})^{2} + (l_{1}m_{2} + l_{2}m_{1})(x – x_{0})(y – y_{0}) + m_{1}m_{2}(y – y_{0})^{2} = 0

⇒ a( x – x_{0})^{2} + 2h( x – x_{0})( y – y_{0}) + b( y – y_{0})2^{2} = 0

Theorem:

The equation to the pair of lines passing through the origin and perpendicular to ax^{2} + 2hxy + by^{2} = 0 is bx^{2} – 2hxy + ay^{2} = 0 .

Proof:

Let ax^{2} + 2hxy + by^{2} = 0 represent the lines l_{1}x + m_{1}y = 0 ………(1) and l_{2}x + m_{2}y = 0 ……..(2).

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

The equation of the line perpendicular to (1) and passing through the origin is m_{1}x – l_{1}y = 0 ……….(3)

The equation of the line perpendicular to (2) and passing through the origin is m_{2} x – l_{2} y = 0 — (4)

The combined equation of (3) and (4) is

⇒ (m_{1}x – l_{1} y)(m_{2} x – l_{2} y) = 0

⇒ m_{1}m_{2}x – (l_{1}m_{2} + l_{2}m_{1} )ny + l_{1}l_{2} y = 0

bx^{2} – 2hxy + ay^{2} = 0

Theorem:

The equation to the lines passing through (x_{0}, y_{0}) and Perpendicular to ax^{2} + 2hxy + by^{2} = 0 is b(x – x_{0})^{2} – 2h(x – x_{0})(y – y_{0}) + a(y – y_{0})^{2} = 0

**Area of the triangle:**

Theorem:

The area of triangle formed by the lines ax^{2} + 2hxy + by^{2} = 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 ax^{2} + 2hxy + by^{2} = 0 represent the lines l_{1}x + m_{1}y = 0 ………(1) and l_{2}x + m_{2}y = 0 ……..(2).

Then l_{1}l_{2} = a, l_{1}m_{2} + l_{2}m_{1} = 2h , m_{1}m_{2} = 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 ax^{2} + 2hxy + by^{2} = 0 is \(\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

Proof:

Let ax^{2} + 2hxy + by^{2} = 0 represent the lines l_{1}x + m_{1}y = 0 — (1) and l_{2}x + m_{2} y = 0 …………..(2).

Then l_{1}l_{2} = a, l_{1}m_{2} + l_{2}m_{1} = 2h , m_{1}m_{2} = 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 ≡ ax^{2} + 2hxy + by^{2} + 2 gx + 2 fy + c = 0 represents a pair of straight lines then

(i) Δ ≡ abc + 2fgh – af^{2} – bg^{2} – ch^{2} =0 and

(ii) h^{2} ≥ ab, g^{2} ≥ ac, f^{2} ≥ bc

Proof:

Let the equation S = 0 represent the two lines l_{1}x + m_{1} y + n_{1} = 0 and l_{2} x + m_{2} y + n_{2} = 0. Then

ax2 + 2hxy + by2 + 2 gx + 2 fy + c

≡ (l_{1}x + m_{1}y + n_{1})(l_{2} x + m_{2} y + n_{2}) = 0

Equating the co-efficients of like terms, we get

l_{1}l_{2} = a, l_{1}m_{2} + l_{2}m_{1} = 2h , m_{1}m_{2} = b, and l_{1}n_{2} + l_{2}n_{1} = 2g , m_{1}n_{2} + m_{2}n_{1} = 2 f , n_{1}n_{2} = c

(i) Consider the product(2h)(2g)(2f)

= (l_{1}m_{2} + l_{2}m_{1})(l_{1}n_{2} + l_{2}n_{1})(m_{1}n_{2} + m_{2}n_{1})

= l_{1}l_{2} (m_{1}^{2}n_{2} + m_{2}^{2}n_{1}^{2}) + m_{1}m_{2} (l_{1}^{2}n_{2}^{2} + l_{2}^{2}n_{1}^{2}) + n_{1}n_{2} (l_{1}^{2}m_{2}^{2} + l_{2}^{2}m_{1}^{2}) + 2l_{1}l_{2}m_{1}m_{2}n_{1}n_{2}

= l_{1}l_{2}[(m_{1}n_{2} + m_{2}n_{1}) – 2m_{1}m_{2}n_{1}n_{2}] + m_{1}m_{2}[(l_{1}n_{2} + l_{2}n_{1}) – 2l_{1}l_{2}n_{1}n_{2}] + n_{1}n_{2}[(l_{1}m_{2} + l_{2}m_{1}) – 2l_{1}l_{2}m_{1}m_{2}] + 2l_{1}l_{2}m_{1}m_{2}n_{1}n_{2}

= a(4 f^{2} – 2bc) + b(4g^{2} – 2ac) + c(4h^{2} – 2ab)

8 fgh = 4[af^{2} + bg^{2} + ch^{2} – abc]

abc + 2 fgh – af^{2} – bg^{2} – ch^{2} = 0

(ii) h^{2} – 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 g^{2} > ac and f^{2} ≥ bc

Note :

If A = abc + 2 fgh – af^{2} – bg^{2} – ch^{2} = 0 , h^{2} ≥ ab, g^{2} ≥ ac and f^{2} ≥ bc, then the equation S ≡ ax^{2} + 2hxy + by^{2} + 2 gx + 2 fy + c = 0 represents a pair of straight lines

Conditions For Parallel Lines-Distance Between Them:

Theorem:

If S = ax^{2} + 2hxy + by^{2} + 2 gx + 2 fy + c = 0 represents a pair of parallel lines then h^{2} = ab and bg^{2} = af^{2}. 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 + n_{1} = 0 ……….(1) lx + my + n_{2} = 0 ………..(2)

ax^{2} + 2hxy + 2gx + 2 fy + c

= (lx + my + n_{1})(lx + my + n_{2})

Equating the like terms

l^{2} = a ………(3)

2lm = 2h …………(4)

m^{2} = b ………..(5)

l(n_{1} + n_{2}) = 2g …….(6)

m(n_{1} + n_{2}) = 2 f ….(7)

n_{1}n_{2} = c …….(8)

From (3) and (5), l^{2}m^{2} = ab and from (4) h^{2} = ab .

**Point of Intersection of Pair of Lines:**

Theorem:

The point of intersection of the pair of lines represented by

a^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 when h^{2} > 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 (x_{1}, y_{1}).

Transfer the origin to (x_{1}, y_{1}) without changing the direction of the axes.

Let (X, Y) represent the new coordinates of (x, y). Then x = X + x_{1} and y = Y + y_{1}.

Now the given equation referred to new axes will be

a( X + x_{1})2 + 2h( X + x_{1})(Y + y_{1}) +b(Y + y_{1})2 + 2 g (X + x_{1}) + 2 f (Y + y_{1}) + c = 0

⇒ aX^{2} + 2hXY + bY^{2} + 2 X (ax_{1} + hy_{1} + g) + 2Y(hx_{1} + by_{1} + f) +(ax_{1}^{2} + 2hx_{1}y_{1} + by^{2} + 2 gx_{1} + 2 fy_{1} + 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,

ax_{1} + hy_{1} + g = 0

hx_{1} + by_{1} + f = 0

ax_{1}^{2} + 2hx_{1} y_{1} + by_{1}^{2} + 2 gx_{1} + 2 fyx + c = 0

But (3) can be rearranged as

x_{1}(ax_{1} + hy + g) + y (hx_{1} + byx + f) + (gx_{1} + fq + c) = 0

⇒ gx_{1} + fy_{1} + c = 0 ………..(4)

Solving (1) and (2) for x_{1} and y_{1}

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 ax^{2} + 2hxy + by^{2} = 0 and the pair of lines ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 form a rhombus then (a – b) fg+h(f^{2} – g^{2}) = 0.

Proof:

The pair of lines ax^{2} + 2hxy + by^{2} = 0 …………(1) is parallel to the lines ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 ……….. (2)

Now the equation

ax^{2} + 2hxy + by^{2} + 2gx + 2 fy + c + λ(ax^{2} + 2hxy + by^{2}) = 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

⇒ g^{2}h – afg = hf^{2} – bfg

⇒ (a – b)fg + h(f^{2} – g^{2}) = 0

\(\frac{g^{2}-f^{2}}{a-b}=\frac{f g}{h}\)

Theorem:

If ax^{2} + 2hxy + by^{2} = 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(x_{1}, y_{1}) and Q(x_{2}, y_{2}) 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 ax^{2} + 2hxy+by^{2} = 0

and px + qy = 1 ………..(2)

ax^{2} + 2hx\(\left(\frac{1-p x}{q}\right)\) + b\(\left(\frac{1-p x}{q}\right)^{2}\) = 0

⇒ x^{2} (aq^{2} – 2hpq + bp^{2}) + 2 x(hp – bp) + b = 0

The roots of this quadratic equation are x_{1} and x_{2} where

x_{1} + x_{2} = \(-\frac{2(h q-b p)}{a q^{2}-2 h p q-b p^{2}}\)

⇒ 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 y_{1},

y_{2} are its roots where

y_{1} + y_{2} = \(-\frac{2(h p-a q)}{a q^{2}-2 h p q-n p^{2}}\) ⇒ 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).