Question

Find the equation to the locus of a point which is always equidistant from the points whose coordinates are $(a+b,a-b)$ and $(a-b,a+b)$

Answer 1

Let the point be $P(h,k)$

$A(a+b,a-b)$ and $B(a-b,a+b)$

$PA=\sqrt{{(h-(a+b\left)\right)}^2+{(k-(a-b\left)\right)}^2}PB=\sqrt{{(h-(a-b\left)\right)}^2+{(k-(a+b\left)\right)}^2}$

Given $PA=PB$

$\sqrt{{(h-(a+b\left)\right)}^2+{(k-(a-b\left)\right)}^2}=\sqrt{{(h-(a-b\left)\right)}^2+{(k-(a+b\left)\right)}^2}{(h-(a+b\left)\right)}^2+{(k-(a-b\left)\right)}^2={(h+(a-b\left)\right)}^2+{(k-(a+b\left)\right)}^2{h}^2+{(a+b)}^2-2h(a+b)+k^2+{(a-b)}^2-2k(a-b)=h^2+{(a-b)}^2-2h(a-b)+k^2+{(a+b)}^2-2k(a+b)-2ah-2bh-2ak+2kb=-2ah+2hb-2ak-2kb-4bh=-4kbh=k$

Replacing $h$ by $x$ and $k$ by $y$

$x=y$

is the required equation of locus.