Question

A$(1,3)$, B$(3,7)$ & $C(7,15)$ are three points. $P$ is the mid point of $AB,Q$ is the mid point of $BC$. Locus of a point $R$ which satisfies $(PR{)}^2-(QR{)}^2=(AC{)}^2$ is

• A. $x+y=50$
• B. $2x^2+2y^2-14x-32y=5$
• C. $6x+12y=297$
• D. none of these

Answer 1

The correct option is C $6x+12y=297$

Coordinates of $P$ are $(\frac{3+1}{2},\frac{7+3}{2})=(2,5)$
and of $Q$ are $(\frac{3+7}{2},\frac{7+15}{2})=(5,11)$.
Let the coordinates of $R$ be $(x,y)$, the
$(PR{)}^2-(QR{)}^2=(AC{)}^2$
$\Rightarrow (x-2{)}^2+(y-5{)}^2-\left[\right(x-5{)}^2+(y-11{)}^2]=\left[\right(7-1{)}^2+(15-3{)}^2]$
$\Rightarrow 6x+12y+4-121=36+144$
$\Rightarrow 6x+12y=297$ which is the locus of $R$.