Question

$(2,3)$ is a point on the side $AB$ of $△ABC.$ The third vertex $C$ moves such that the sides $AC,BC$ are bisected by $x^2-y^2=0$ at right angles. Then $C$ lies on

• A. $2x-3y=0$
• B. $3x-2y=0$
• C. $2x+3y=0$
• D. $3x+2y=0$

Answer 1

The correct option is A $2x-3y=0$


$x^2-y^2=0$
$\Rightarrow (x-y)(x+y)=0$
$\Rightarrow x-y=0,x+y=0$
Let $C\equiv (h,k)$
Then $A,B$ are the images of $C$ in $x-y=0,x+y=0$
$\therefore A\equiv (k,h),B\equiv (-k,-h)$
Slope of $AB$ is $\frac{h}{k}=\frac{h-3}{k-2}$
$\Rightarrow 2h-3k=0$