Let $A, B$ and $C$ be three points in a plane. The locus of a point $P$ moving such that $P A^2+P B^2=2 P…

Let $A, B$ and $C$ be three points in a plane. The locus of a point $P$ moving such that $P A^2+P B^2=2 P C^2$ is a
  1. Straight line
  2. Pair of straight lines
  3. Circle
  4. Parabola

Solution

Let $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ and $p(x, y)$ be points on $x y$ plane. Now, $ P A^2+P B^2=2 P C^2 $ $ \begin{gathered} \Rightarrow\left(x-x_1\right)^2+\left(y-y_1\right)^2+\left(x-x_2\right)^2+\left(y-y_2\right)^2 \\ =2\left[\left(x-x_3\right)^2+\left(y-y_3\right)^2\right] \\ \Rightarrow x^2+x_1^2-2 x x_1+y^2+y_1^2-2 y y_1 \\ \quad+x^2+x_2^2-2 x x_2+y^2+y_2^2-2 y y_2 \\ =2\left[x^2+x_3^2-2 x x_3+y^2+y_3^2-2 y y_3\right] \\ \Rightarrow x\left(-2 x_1-2 x_2+4 x_3\right)+y\left(-2 y_1-2 y_2+4 y_3\right) \\ \quad+x_1^2+x_2^2+y_1^2+y^2-2 x_3^2-2 y_3^2=0 \end{gathered} $ this equation is in the form of $ a x+b y+c=0 $ Hence, locus of $p$ is a straight line

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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