Question

Let $A=(1,0)$ , $B=(-1,0),C=(2,0)$, the locus of a point $P$ such that $P{B}^2+P{C}^2=2P{A}^2$ is

• A. a straight line parallel to $x$-axis
• B. a straight line parallel to $y$-axis
• C. pair of straight line
• D. combined equation of coordiante axes

Answer 1

The correct option is B a straight line parallel to $y$-axis

Let $P=(x,y)$
Then, $PA=\sqrt{(x-1{)}^2+y^2}$ ...(i)
$PB=\sqrt{(x+1{)}^2+y^2}$ ...(ii)
$PC=\sqrt{(x-2{)}^2+y^2}$ ...(iii)
Hence, $P{C}^2+P{B}^2=2P{A}^2$
$\Rightarrow \left[\right(x-2{)}^2+y^2]+[(x+1{)}^2+y^2]=2(x-1{)}^2+2y^2$
$\Rightarrow (x-2{)}^2+(x+1{)}^2+2y^2=2(x-1{)}^2+2y^2$
$\Rightarrow (x-2{)}^2+(x+1{)}^2=2(x-1{)}^2$
$\Rightarrow (x^2-4x+4)+(x^2+2x+1)=2(x^2-2x+1)$
$\Rightarrow 2x^2-2x+5=2x^2-4x+2$
$\Rightarrow 2x+3=0$
$\Rightarrow x=\frac{-3}{2}$
Hence, this represents a line parallel to $y$-axis.