Question

The area (in sq. units) of the triangle formed by the straight line $x+y=3$ and the angular bisectors of the pair of straight lines $x^2-y^2+2y=1$, is

• A. 1
• B. 2
• C. 3
• D. 6

Answer 1

Answer: (B)

pair of straight line is $x^2-y^2+2y=1$
$\Rightarrow x^2-\left(y^2-2y+1\right)=0$
$\Rightarrow x^2-(y-1{)}^2=0$
$\Rightarrow (x-y+1)(x+y-1)=0$
So, lines are $x+y-1=0$ and $x-y+1=0$
equation of angle bisector is
$\frac{A_{1}x+B_{1}y+c_1}{\sqrt{A_1^2+B_1^2}}=\pm \frac{A_{2}x+B_{2}y+c_2}{\sqrt{A_2^2+B_2^2}}$
$\Rightarrow \frac{x+y-1}{\sqrt{2}}=\pm \frac{x-y+1}{\sqrt{2}}$
So, $\frac{x+y-1}{\sqrt{2}}=\frac{x-y+1}{\sqrt{2}}$
and $\frac{x+y-1}{\sqrt{2}}=\frac{-(x-y+1)}{\sqrt{2}}$
$\Rightarrow x+y-1=x-y+1$

and
$x+y-1=-x+y-1$
$\Rightarrow 2y=2$ and $x=0$
$\Rightarrow y-1=0$
$\Rightarrow y=1$
Now, vertices of $\Delta$ formed by $x=0,y=1$ and
$x+y=3$ are $(0,1),(0,3)$ and $(2,1)$
Area of $\Delta =\frac{1}{2}(2)(2)=2$ s q . units.