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Question

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a. A straight line parallel to x-axis

b. Circle through the origin

c. Circle with center through the origin

d. A straight line parallel to y-axis

Answer
Verified

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.

Complete step-by-step answer:

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.

Complete step-by-step answer:

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.