Question

Find a point on $y-axis$ which is equidistant from $(-5,-2)$ and $(3,2)$.

Answer 1

Let the point $A$ on y-axis be $(0,y)$

Given that point $A$ is equidistant from $(-5,-2)and(3,2)$

Hence, using distance formula we have

$\sqrt{(0-(-5){)}^2+(y-(-2){)}^2}=\sqrt{(0-3{)}^2+(y-2{)}^2}$

$\sqrt{25+(y+2{)}^2}=\sqrt{9+(y-2{)}^2}$

squaring both sides we get

$25+(y^2+4+4y)=9+(y^2+4-4y)$

$4y+4y=y^2+4+9-y^2-4-25$

$8y=-16$

$y=\frac{-16}{8}=-2$

Hence, the point $A$ on y-axis is $(0,-2)$