Question

The distance of $P(x,y)$ from $A(5,1)$ and $V(-1,5)$ are equal then $\frac{x}{y}=\frac{p}{q}$. Find the value of $p+q$

Answer 1

Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ can be calculated using the formula $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
Given,

Distance between the points $(x,y);(5,1)=$ Distance between $(x,y);(-1,5)$

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

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

$\Rightarrow 5^2+x^2-10x+1^2+y^2-2y=1^2+x^2+2x+5^2+y^2-10y$

$\Rightarrow -10x-2y=+2x-10y$

$\Rightarrow 12x=8y$

$\Rightarrow \frac{x}{y}=\frac{2}{3}$

Hence, $p+q=5$.