Question

The locus of the point which is equidistant from the points $A(2,3)$ and $B(-3,4)$ is

• A. $5x-y+6=0$
• B. $5x+y+6=0$
• C. $x+5y+6=0$
• D. $x-5y+6=0$

Answer 1

The correct option is A $5x-y+6=0$

Let the equidistant point be $P(h,k)$
$PA=PB$
$\Rightarrow P{A}^2=P{B}^2\Rightarrow (h-2{)}^2+(k-3{)}^2=(h+3{)}^2+(k-4{)}^2\Rightarrow 4-4h+9-6k=9+6h+16-8k\Rightarrow 5h-k+6=0$

Hence, the locus of $P$ is $5x-y+6=0$