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If the points $\mathrm{P}(\mathrm{x}, \mathrm{y})$ is point equidistant from the points $\mathrm{A}(5,1)$ and $\mathrm{B}(-1,5)$, Prove that 3x=2y.
If the points $\mathrm{P}(\mathrm{x}, \mathrm{y})$ is point equidistant from the points $\mathrm{A}(5,1)$ and $\mathrm{B}(-1,5)$, Prove that 3x=2y.
CBSE | Class 10 | Coordinate Geometry | Exercise 16.1 | Maths | RS Aggarwal
If the points $\mathrm{P}(\mathrm{x}, \mathrm{y})$ is point equidistant from the points $\mathrm{A}(5,1)$ and $\mathrm{B}(-1,5)$, Prove that 3x=2y.

$\mathrm{AP}=\mathrm{BP}$

$\Rightarrow \sqrt{(x-5)^{2}+(y-1)^{2}}=\sqrt{(x+1)^{2}+(y-5)^{2}}$

$\Rightarrow(x-5)^{2}+(y-1)^{2}=(x+1)^{2}+(y-5)^{2}$

$\text { (Squaring both sides) }$

$\Rightarrow x^{2}-10 x+25+y^{2}-2 y+1=x^{2}+2 x+1+y^{2}-10 y+25$

 

$$
\begin{aligned}
&\Rightarrow-10 x-2 y=2 x-10 y \\
&\Rightarrow 8 y=12 x \\
&\Rightarrow 3 x=2 y \\
&\text {Therefore, } 3 x=2 y
\end{aligned}
$$

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