Question

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

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is A $2$

The given lines are $x^2-{(y-1)}^2=0$ or $x-y+1=0$, $x+y-1=0$
Their bisecots are $x=0,y=1$. The sides of the triangle are $x=0,y=1$ and $x+y=3$
Solving in pairs the vertices of $△ABC$ are $(0,1),(0,3)$ and $(2,1)$. One side along y-axis
Area $=\frac{1}{2}b\times h=\frac{1}{2}2.2=2$