First slide

Pair of straight lines

Question

The equation of straight line bisecting the acute angle between the lines represented by the equation $7 x^{2} + 6 x y - y^{2} - 5 x + 3 y - 2 = 0$

Moderate

A

$3 x - y + 1 = 0$
B

$2 x + 6 y - 7 = 0$
C

$12 x + 4 y - 3 = 0$
D

$2 x - 6 y + 7 = 0$

Solution

$7 x^{2} + 6 x y - y^{2} - 5 x + 3 y - 2 = (7 x - y + n_{1}) (x + y + n_{2})$
coefficient of x is $- 5 = n_{1} + 7 n_{2}$
coefficient of y is $3 = n_{1} - n_{2}$
solving we get $n_{1} = 2, n_{2} = - 1$ lines are
$7 x - y + 2 = 0, x + y - 1 = 0 \Rightarrow - x - y + 1 = 0$
$a_{1} = 7, b_{1} = - 1, c_{1} = 2, a_{2} = - 1, b_{2} = - 1, c_{2} = 1$
$c_{1} c_{2} > 0, a_{1} a_{2} + b_{1} b_{2} < 0$
acute angle bisector is $\frac{a_{1} x + b_{1} y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}} = + \frac{a_{2} x + b_{2} y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}$
$\Rightarrow 12 x + 4 y - 3 = 0$

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