Question

The area bounded by the angle bisectors of the lines $x^2-y^2+2y=1$ and the line $x+y=3$, is

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is A $2$

Area bounded by angle bisectors of line $x^2-y^2+2y=1$ and the line $x+y=3$

Angle bisector of line

$x^2-(y^2+1-2y)=0x^2-(y-1{)}^2=0x-y+1=0x+y-1=0\left|\frac{x-y+1}{\sqrt{1^2+(-1{)}^2}}\right|=\left|\frac{x+t-1}{\sqrt{1^2+1^2}}\right|\left|\frac{x-y+1}{\sqrt{2}}\right|=\left|\frac{x+t-1}{\sqrt{2}}\right|x-y+1=x+y-1y=1x-y+1=-x-y+1x=0$

Area bounded by lines $x=0,y=1,x+y=3$ is

Area$=\frac{1}{2}\times l\times b=\frac{1}{2}\times 2\times 2$

Hence the area comes to be equal to $2sq.units$