Question

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

• A. $2$ sq. units
• B. $4$ sq. units
• C. $3$ sq. units
• D. $8$ sq. units

Answer 1

The correct option is A $2$ sq. units

$x^2-y^2+2y=1$
$\Rightarrow x^2-(y-1{)}^2=0$
$\Rightarrow (x-y+1)(x+y-1)=0$
Bisectors of above lines are $x=0$ and $y=1$


So, area between $x=0,y=1$ and $x+y=3$ is $\frac{1}{2}\times 2\times 2=2$ sq. units