If $P = (1, 0), Q = (- 1, 0), R = (2, 0)$ are $3$ given points, then the locus of the point S satisfying the relation $S Q^{2} + S R^{2} = 2 S P^{2}$ is

a straight line parallel to the x-axis

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circle through the origin

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circle with centre at the origin

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a straight line parallel to the y-axis.

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Solution

The correct option is D a straight line parallel to the y-axis.
Given points
$P (1, 0), Q (- 1, 0), R (2, 0)$
Let point $S (x, y)$
$S Q^{2} + S R^{2} = 2 S P^{2}$
$(x + 1)^{2} + y^{2} + (x - 2)^{2} + y^{2} = 2 ((x - 1)^{2} + y^{2})$
$x^{2} + 1 + 2 x + y^{2} + x^{2} + 4 - 4 x + y^{2} = 2 (x^{2} + 1 - 2 x + y^{2})$
$2 x^{2} - 2 x + 2 y^{2} + 5 = 2 x^{2} + 2 - 4 x + 2 y^{2}$
$- 2 x + 5 = 2 - 4 x$
$2 x = - 3 \Rightarrow x = - \frac{3}{2}$
Line $x = - \frac{3}{2}$ represents a line which is parallel to y-axis

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