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Q.
Find the area of the triangle formed by the line $x+y=3$ and the angle bisectors of pair of straight lines $x^2-y^2+2 y=1$
Straight Lines
Solution:
$ x^2-(y-1)^2=0$
$x-y+1=0, x+y-1=0$
$\Rightarrow$ Bisector are $y$-axis and line parallel to $x$-axis, i.e. $y =1$.
Area of triangle $=\frac{1}{2} \cdot 2 \times 2=2$ sq. units
