Find the relation between $x$ and $y$ such that the point $P(x, y)$ is equidistant from the points $A(1, 4)$ and $B(-1, 2)$.

Let $P\left(x,y\right)$ be equidistance from the points $A\left(1,4\right)$ and $B\left(−1,2\right)$

Given:$AP=BP$

${AP}^{2}={BP}^{2}$

By distance formula,

${\left(x-1\right)}^{2}+{\left(y-4\right)}^{2}={\left(x+1\right)}^{2}+{\left(y-2\right)}^{2}$

$\Rightarrow\,{x}^{2}-2x+1+{y}^{2}-8y+16={x}^{2}+2x+1+{y}^{2}-4y+4$

$\Rightarrow\,-2x+1-8y+16-2x-1+4y-4=0$

$\Rightarrow\,-4x-4y+12=0$

$\Rightarrow\,x+y-3=0$ gives the relation between $x$ and $y$.