Question

Area of the triangle formed by the line $x+y=3$ and the angle bisectors of the pairs of straight lines $x^2-y^2+2y=1$ is?

Answer 1

$x^2-y^2+2y=1$

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

Hence, the two lines are =

$y=-x+1$ and $y=x+1$

Equation of angle bisectors of these lines

$x=0$ and $y=1$

Intersection between the 3 lines is

$A(0,3),B(0,1),C(2,1)$

It can be seen that as the points A and B lie on y-axis only and BC is parallel to x-axis. Hence, $△ABC$ is a right angled triangle.

$Area=\frac{1}{2}\times AB\times BC$

$⟹\frac{1}{2}\times 2\times 2$

$⟹2$