Question

Find the equation of locus of P, if the line segment joining $(2,3)$ and $(-1,5)$ subtends a right angle at P.

Answer 1

Let coordinates of $P$ be $(h,k)$.

Slope of line joining $P$ and $(2,3)$ = $m_1=\frac{k-3}{h-2}$

Slope of line joining $P$ and $(-1,5)$ = $m_2=\frac{k-5}{h+1}$

Now, as a right angle is subtended at $P$,

$m_1=-\frac{1}{m_2}$

$\therefore \frac{k-3}{h-2}=-\frac{h+1}{k-5}$

$\therefore (k-3)(k-5)+(h+1)(h-2)=0$

$\therefore k^2-8k+15+h^2-h-2=0$

$\therefore h^2+k^2-h-8k+13=0$

$\therefore h^2-h+0.25+k^2-8k+16+13-16.25=0$

$\therefore (h-0.5{)}^2+(k-4{)}^2=3.25$

Hence, the required locus is $(x-0.5{)}^2+(y-4{)}^2=3.25$.