Question

Find a point on the y-axis which is equidistant from the points $(5,4)$ and $(-2,3)$.

Answer 1

The point on the y-axis will be of the form of $(0,y)$

This point is equidistant from the point $(x_1,y_1)=(5,4)\&(x_2,y_2)=(-2,3)$

We know that distance between two points is $\mathrm{S}=\sqrt{{({\mathrm{x}}_2-{\mathrm{x}}_1)}^2+{({\mathrm{y}}_2-{\mathrm{y}}_1)}^2}$

So, the distance of given points from point $(0,\mathrm{y})$ must be equal

i.e. $\sqrt{{(0-5)}^2+{(\mathrm{y}-4)}^2}=\sqrt{{\left\{0-(-2)\right\}}^2+{(\mathrm{y}-3)}^2}$

Squaring on both sides, we get

$25+{(\mathrm{y}-4)}^2=4+{(\mathrm{y}-3)}^2$

$\Rightarrow$$25+{\mathrm{y}}^2-8\mathrm{y}+16=4+{\mathrm{y}}^2-6\mathrm{y}+9$

$\Rightarrow -2\mathrm{y}=-28$

$\mathrm{y}=14$

Hence the point is $(0,14)$ which is equidistant from points $(5,4)$ & $(-2,3)$.