Question

Find a relation between $x$ and $y$ such that the point $(x,y)$ is equidistant from the point $(3,6)$ and $(-3,4)$.

Answer 1

Point $(x,y)$ is equidistant from $(3,6)$ and $(-3,4)$.
And Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
$\therefore \sqrt{(x-3{)}^2+(y-6{)}^2}=\sqrt{(x-(-3){)}^2+(y-4{)}^2}$
$\Rightarrow \sqrt{(x-3{)}^2+(y-6{)}^2}=\sqrt{(x+3{)}^2+(y-4{)}^2}$
$\Rightarrow (x-3{)}^2+(y-6{)}^2=(x+3{)}^2+(y-4{)}^2$
$\Rightarrow x^2+9-6x+y^2+36-12y=x^2+9+6x+y^2+16-8y$
$\Rightarrow 36-16=6x+6x+12y-8y$
$\Rightarrow 20=12x+4y$
$\Rightarrow 12x+4y=20$
$\Rightarrow 3x+y=5$
$\Rightarrow 3x+y-5=0$