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

Let the point $P(x,y)$ is equidistant from the points $A(3,6)$and $B(-3,4)$

$\therefore PA=PB$

By using the distance formula

$=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$ we have

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

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

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

$\Rightarrow -6x-6x-12y+8y+36-16=0$

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

$\Rightarrow 3x+y-5=0$

Hence this is a relation between $x$ and $y$.