Question

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

Answer 1

Given: $P(x,y),A(3,6),B(-3,4)$

$PA=PB$

$P{A}^2=P{B}^2$

Distance between two points $=\sqrt{{{(x}_2-x_1{)}^2+(y}_2-y_1{)}^2}$

$\because PA=PB$

$\sqrt{(x-3{)}^2+(y-6{)}^2}$ = $\sqrt{(x+3{)}^2+(y-4{)}^2}$

Squaring and expanding both sides,

$\Rightarrow$$x^2+9-6x+y^2+36-12y=x^2+9+6x+y^2+16-8y$

$\Rightarrow$$36-6x-12y=6x-8y+16$

$\Rightarrow$$12x+4y-20=0$

$\Rightarrow$$3x+y=5$