Question

# If $P = (1,0)$, $Q = ( - 1,0)$, $R = (2,0)$ are 3 given point, then the locus of point S satisfying the relationship $S{Q^2} + S{R^2} = 2S{P^2}$ isa. A straight line parallel to x-axisb. Circle through the originc. Circle with center through the origind. A straight line parallel to y-axis

Hint: Here, we need to find the locus of point S satisfying the given equation $S{Q^2} + S{R^2} = 2S{P^2}$ and we need to state the relation whether it is a straight line or circle.

We need to find the locus of the points S satisfying the relation $S{Q^2} + S{R^2} = 2S{P^2} \to (1)$
Let point S have coordinates $(x,y)$. Coordinates of P is $(1,0)$, Q is $( - 1,0)$ and R is $(2,0)$.
Now using the distance formulae $SP = \sqrt {{{(x - 1)}^2} + {{(y - 0)}^2}}$ and $SQ = \sqrt {{{(x + 1)}^2} + {{(y - 0)}^2}}$ and $SR = \sqrt {{{(x - 2)}^2} + {{(y - 0)}^2}}$.
Substituting the above in equation (1)
We have
${(x + 1)^2} + {(y - 0)^2} + {(x - 2)^2} + {(y - 0)^2} = 2({(x - 1)^2} + {(y - 0)^2})$
Simplifying if we get using ${(a + b)^2} = {a^2} + 2ab + {b^2}$ and ${(a - b)^2} = {a^2} - 2ab + {b^2}$
${x^2} + 2x + 1 + {y^2} + {x^2} - 4x + 4 + {y^2} = 2({x^2} - 2x + 1 + {y^2})$
Simplifying further,
${x^2} + 2x + 1 + {y^2} + {x^2} - 4x + 4 + {y^2} = 2{x^2} - 4x + 2 + 2{y^2}$
On solving, we get,
$2x + 3 = 0 \\ {\text{or x = }}\dfrac{{ - 3}}{2} \\$
Clearly $x = \dfrac{{ - 3}}{2}$ is a straight line in the second & third quadrants which is parallel to y axis hence (d) is the right option.

Note: Locus refers to the family of curves, so whenever we need to find the locus that is a family of curves satisfying a specific equation then simply solve and simplify to obtain the final relation between x and y to obtain locus.