Equation $2 x^{2} + 5 x y + 3 y^{2} + 6 x + 7 y + 4 = 0$ represents a pair of straight lines. If their point of intersection is (h,k) find the value of h-k.

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Solution

It is given that equation $2 x^{2} + 5 x y + 3 y^{2} + 6 x + 7 y + 4 = 0$ represents the pair of straight lines

We can write this equation as quadratic equation in x

We have,

$2 x^{2} + (5 y + 6) x + 3 y^{2} + 7 y + 4 = 0$ -----------------------(1)

Roots of the given equation

$X = \frac{- (5 y + 6) \pm \sqrt{(5 y + 6)^{2}} - 4.2 (3 y^{2} + 7 y + 4)}{4}$

$= \frac{- (5 y + 6) \pm \sqrt{25 y^{2} + 60 y + 36 - 24 y^{2} - 56 y - 32}}{4}$

$X = \frac{- (5 y + 6) \pm \sqrt{y^{2} + 4 y + 4}}{4}$

$X = \frac{- (5 y + 6) \pm (y + 2)}{4}$

$X = \frac{- 5 y - 6 + y + 2}{4}, \frac{- 5 y - 6 - y - 2}{4}$

Or 4x = -4y - 4 , 4x + 6y + 8 = 0

X + y + 1 = 0 and 2x + 3y + 4 = 0

Hence, equation 1 represents the pair of straight lines whose equations are

X + y + 1 = 0------------------2

2x + 3y + 4= 0-----------------3

Multiplying 2 in equation 2 and subtracting from equation 3

2x + 2y + 2 = 0

2x + 3y + 4 = 0

-y - 2 = 0

-y = 2

y = -2

Substituting in equation 1

-2+x+1=0

X=1

Point of intersection (1,-2)

h=1, k=-2

h-k= 1-(-2)=3

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