Question

If P(1, 0), Q(– 1, 0) and R(2, 0) are three given points. The point S satisfies the relation \[S{{Q}^{2}}+S{{R}^{2}}=2S{{P}^{2}}.\] The locus of S meets PQ at the point
\[\left( a \right)\left( 0,0 \right)\]
\[\left( b \right)\left( \dfrac{2}{3},0 \right)\]
\[\left( c \right)\left( -\dfrac{3}{2},0 \right)\]
\[\left( d \right)\left( 0,-\dfrac{2}{3} \right)\]

Answer 1

Hint: First, let the coordinate of S be (x, y). Now it is given that the point S satisfies \[S{{Q}^{2}}+S{{R}^{2}}=2S{{P}^{2}},\] so we will find the distance SQ, SR, SP using the distance formula, i.e \[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}.\] Then, we will substitute in this equation to find the equation of the locus. Once we have the equation of the locus S, we will find the point where it touches PQ.

Complete step-by-step answer:
Let the point S have coordinates as (x, y). First of all, we will find the locus of S and then find the point. Now as the first step, we will find the distance between each set of points as below,
SQ = Distance between S and Q
We know that the distance formula is given by \[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}.\]
Here we have S(x, y) and Q(– 1, 0). Therefore we get,
\[SQ=\sqrt{{{\left( x-\left( -1 \right) \right)}^{2}}+{{\left( y-0 \right)}^{2}}}\]
\[\Rightarrow SQ=\sqrt{{{\left( x+1 \right)}^{2}}+{{y}^{2}}}\]
By squaring both the sides, we get,
\[\Rightarrow S{{Q}^{2}}={{\left( x+1 \right)}^{2}}+{{y}^{2}}\]
Now, SR is the distance between S and R. Here we have S(x, y) and R(2, 0). Therefore, applying the distance formula again, we get,
\[SR=\sqrt{{{\left( x-2 \right)}^{2}}+{{\left( y-0 \right)}^{2}}}\]
\[\Rightarrow SR=\sqrt{{{\left( x-2 \right)}^{2}}+{{y}^{2}}}\]
By squaring both the sides, we get,
\[\Rightarrow S{{R}^{2}}={{\left( x-2 \right)}^{2}}+{{y}^{2}}\]
Now, SP is the distance between S and P. Here we have S(x, y) and P(1, 0). Therefore, applying the distance formula again, we get,
\[SP=\sqrt{{{\left( x-1 \right)}^{2}}+{{\left( y-0 \right)}^{2}}}\]
\[\Rightarrow SP=\sqrt{{{\left( x-1 \right)}^{2}}+{{y}^{2}}}\]
By squaring both the sides, we get,
\[\Rightarrow S{{P}^{2}}={{\left( x-1 \right)}^{2}}+{{y}^{2}}\]
Now using this value of \[S{{P}^{2}},S{{Q}^{2}},S{{R}^{2}}\] in \[S{{Q}^{2}}+S{{R}^{2}}=2S{{P}^{2}},\] we get,
\[{{\left( x+1 \right)}^{2}}+{{y}^{2}}+{{\left( x-2 \right)}^{2}}+{{y}^{2}}=2{{\left( x-1 \right)}^{2}}+2{{y}^{2}}\]
By simplifying the above terms, we get,
\[\Rightarrow {{\left( x+1 \right)}^{2}}+{{\left( x-2 \right)}^{2}}+2{{y}^{2}}=2{{\left( x-1 \right)}^{2}}+2{{y}^{2}}\]
Cancelling out \[2{{y}^{2}},\] we get,
\[\Rightarrow {{\left( x+1 \right)}^{2}}+{{\left( x-2 \right)}^{2}}=2{{\left( x-1 \right)}^{2}}\]
Solving the above equation further, we get,
\[\Rightarrow {{x}^{2}}+1+2x+{{x}^{2}}+4-4x=2{{x}^{2}}+2-4x\]
Cancelling out the similar terms, we get,
\[\Rightarrow 2x+3=0\]
Solving further, we get,
\[\Rightarrow 2x=-3\]
Dividing both the sides by 2, we get,
\[x=\dfrac{-3}{2}\]
So, the locus of S is a line \[x=\dfrac{-3}{2}\] parallel to the y-axis. Now PQ is a line joining P (1, 0) and Q (– 1, 0) while the locus of S is the line \[x=\dfrac{-3}{2}.\]
So, locus of S joining PQ at the point \[\left( \dfrac{-3}{2},0 \right).\]
Therefore, the correct option is (c).

Note:While solving for distance, students have to be very clear that the distance between S(x, y) and Q(– 1, 0) is not \[SQ=\sqrt{{{\left( x-1 \right)}^{2}}+{{\left( y-0 \right)}^{2}}}\] it is \[SQ=\sqrt{{{\left( x-\left( -1 \right) \right)}^{2}}+{{\left( y-0 \right)}^{2}}}.\] Similarly, while opening the brackets be very cautious, i.e. \[{{\left( a+b \right)}^{2}}\ne {{a}^{2}}+{{b}^{2}}\] it is \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}.\]