The combined equation of the two lines $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ and $\mathrm{a}^{\prime} \mathrm{x}+\mathrm{b}^{\prime} \mathrm{y}+\mathrm{c}^{\prime}=0$ can be written as $(\mathrm{ax}+\mathrm{by}+\mathrm{c})(\mathrm{a}^{\prime} \mathrm{x}+\mathrm{b}^{\prime} \mathrm{y}+\mathrm{c}^{\prime})=0$ The equation of the angle bisectors of the lines represented by the equation $2 x^{2}+x y-3 y^{2}=0$ is

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Straight Lines

The combined equation of the two lines

ax + by + c = 0

and

a^{'} x + b^{'} y + c^{'} = 0

can be written as

(ax + by + c) (a^{'} x + b^{'} y + c^{'}) = 0

The equation of the angle bisectors of the lines represented by the equation

2 x^{2} + x y - 3 y^{2} = 0

(A) $3 x^{2} + 5 x y + 2 y^{2} = 0$
(B) $x^{2} - y^{2} + 10 x y = 0$
(C) $3 x^{2} + x y - 2 y^{2} = 0$
(D) $x^{2} - y^{2} - 10 x y = 0$ ✓ Correct

Answer: (D)

Solution

For a pair of lines

a x^{2} + 2 h x y + b y^{2} = 0

the bisectors satisfy

\frac{x ^{2} - y ^{2}}{a - b} = \frac{x y}{h}

. Here

a = 2, b = - 3, h = \frac{1}{2}

, so

\frac{x ^{2} - y ^{2}}{5} = \frac{x y}{1/2} = 2 x y

, which gives

x^{2} - y^{2} = 10 x y

. The combined equation of the bisectors is therefore

x^{2} - y^{2} - 10 x y = 0

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Single correct option

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