Question

The locus of a point, which is equidistant from the points $(1,1)$ and $(3,3)$, is

• A. $y=x+4$
• B. $x+y=4$
• C. $x=2$
• D. $y=2$
• E. $y=-x$

Answer 1

The correct option is B $x+y=4$

Let $P(h,k)$ be a point, which is equidistant from the points $A(1,1)$ and $B(3,3)$.
i.e., $PA=PB\Rightarrow (PA{)}^2=(PB{)}^2$ (by distance formula)
$\Rightarrow 1-2h+1-2k=9-6h+9-6k$
$\Rightarrow 4h+4k=16$
$\Rightarrow h+k=4$
So, required locus is $x+y=4$.