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
- $x+y=0$
- $x+y=4$
- $x+y=7$
- $4 x+4 y=38$
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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