The equation of the locus of a point which is equidistant from the points $(2,3)$ and $(4,5)$ is

The equation of the locus of a point which is equidistant from the points $(2,3)$ and $(4,5)$ is
  1. $x+y=0$
  2. $x+y=4$
  3. $x+y=7$
  4. $4 x+4 y=38$

Solution

Let point $P(x, y)$ which is equidistant from the points $(2,3)$ and $(4,5)$. $ \begin{array}{cc} \therefore & (x-2)^2+(y-3)^2=(x-4)^2+(y-5)^2 \\ \Rightarrow & x^2-4 x+4+y^2-6 y+9 \\ \Rightarrow & =x^2-8 x+16+y^2-10 y+25 \\ \Rightarrow & 4 x+4 y=41-13 \\ \Rightarrow & 4(x+y)=28 \\ \Rightarrow & x+y=7 \end{array} $

Asked in: AP EAMCET 2021 (25 Aug Shift 1)

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