Question

If $P=(1,0),Q=(-1,0)$ and $R=(2,0)$ are three given points, then the locus of the point $S$ satisfying the relation ${\left(SQ\right)}^2+{\left(SR\right)}^2=2{\left(SP\right)}^2$, is

• A. A straight line parallel to x-axis
• B. A circle passing through the origin
• C. A circle with the center at the origin
• D. A straight line parallel to the y-axis

Answer 1

The correct option is D

A straight line parallel to the y-axis

Explanation for the correct option

Here, the coordinates of $P,Q,\text{and}R$ are known, Let the coordinates of a point $S$ be $\left(a,b\right)$.

The formula that will be used here is the distance formula which is given by,

$\mathit{d}\mathbf{=}\sqrt{{\mathbf{(}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}{\mathbf{y}}_{\mathbf{2}}\mathbf{-}{\mathbf{y}}_{\mathbf{1}}\mathbf{)}}^{\mathbf{2}}}$, where $d$ is the distance between two points and $\left(x_1,y_1\right)$ are the coordinates of one point and $\left(x_2,y_2\right)$ are the coordinates of another point.

Now, the given equation is ${\left(SQ\right)}^2+{\left(SR\right)}^2=2{\left(SP\right)}^2$, then

$\begin{array}{rcl}{\left(a+1\right)}^2+{\left(b\right)}^2+{\left(a-2\right)}^2+{\left(b\right)}^2& =& 2\left({\left(a-1\right)}^2+{\left(b\right)}^2\right)\\ & \Rightarrow & a^2+1+2a+b^2+a^2+4-4a+b^2=2\left(a^2+1-2a+b^2\right)\\ & \Rightarrow & 2a^2+5-2a+2b^2=2a^2+2-4a+2b^2\\ & \Rightarrow & 5-2a=2-4a\\ & \Rightarrow & -2a=3\\ & \Rightarrow & a=-\frac{3}{2}\end{array}$

Therefore, the line is parallel to $y-\text{axis}$ with intercept on $x=-1·5$.

Hence, the correct option is (D).