Question

$P:x^2-y^2+2y-1=0$
$L:x+y=3$

Area of the triangle formed by the angle bisectors of the pair of lines $P$ and the line $L$ (in square units) is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$P:x^2-{(y-1)}^2=0$

$\Rightarrow x+y-1=0$ and $x-y+1=0$

Equation of the angle bisector is

$\frac{A_{1}x+B_{1}y+C_1}{\sqrt{A_1^2+B_1^2}}=\pm \frac{A_{2}x+B_{2}y+C_2}{\sqrt{A_2^2+B_2^2}}$
$\frac{x+y-1}{\sqrt{2}}=\pm \frac{x-y+1}{\sqrt{2}}$

$\Rightarrow x=0$ or $y-1=0$
Hence vertices of the triangle are $(0,1),(0,3)$ and $(2,1)$

Therefore area of triangle is $2$