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If $p(x, y)$ is point equidistant from the points $A(6,-1)$ and $B(2,3)$, show that $x-y=3$
If $p(x, y)$ is point equidistant from the points $A(6,-1)$ and $B(2,3)$, show that $x-y=3$
CBSE | Class 10 | Coordinate Geometry | Exercise 16.1 | Maths | RS Aggarwal
If $p(x, y)$ is point equidistant from the points $A(6,-1)$ and $B(2,3)$, show that $x-y=3$

points are $A(6,-1)$ and $B(2,3)$. The point $P(x, y)$ equidistant from the points A and B So, PA = PB

$$
\begin{aligned}
&\text { Also, }(P A)^{2}=(P B)^{2} \\
&\Rightarrow(6-x)^{2}+(-1-y)^{2}=(2-x)^{2}+(3-y)^{2} \\
&\Rightarrow x^{2}-12 x+36+y^{2}+2 y+1=x^{2}-4 x+4+y^{2}-6 y+9 \\
&\Rightarrow x^{2}+y^{2}-12 x+2 y+37=x^{2}+y^{2}-4 x-6 y+13 \\
&\Rightarrow x^{2}+y^{2}-12 x+2 y-x^{2}-y^{2}+4 x+6 y=13-37 \\
&\Rightarrow-8 x+8 y=-24 \\
&\Rightarrow-8(x-y)=-24 \\
&\Rightarrow x-y=\frac{-24}{-8} \\
&\Rightarrow x-y=3
\end{aligned}
$$

( Proved)

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