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. $y = 2$

Answer 1

Answer: (B)

$x + y = 4$

Answer 2

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