Question

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

Answer 1

We know that the distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ is

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

Let the given points be $A=(5,2)$ and $B=(-4,3)$ and let the point on y-axis be $P(0,y)$.

We first find the distance between $P(0,y)$ and $A=(5,2)$ as follows:

$PA=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(5-0)}^2+{(2-y)}^2}=\sqrt{5^2+{(2-y)}^2}=\sqrt{25+{(2-y)}^2}$

Similarly, the distance between $P(0,y)$ and $B=(-4,3)$ is:

$PB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(-4-0)}^2+{(3-y)}^2}=\sqrt{{(-4)}^2+{(3-y)}^2}=\sqrt{16+{(3-y)}^2}$

Since the point $P(0,y)$ is equidistant from the points $A=(5,2)$ and $B=(-4,3)$, therefore, $PA=PB$ that is:

$\sqrt{25+{(2-y)}^2}=\sqrt{16+{(3-y)}^2}$

$\Rightarrow (\sqrt{25+{(2-y)}^2}{)}^2=(\sqrt{16+{(3-y)}^2}{)}^2$

$\Rightarrow 25+{(2-y)}^2=16+{(3-y)}^2$

$\Rightarrow {(2-y)}^2-{(3-y)}^2=16-25$

$\Rightarrow {(4+y}^2-4y)-{(9+y}^2-6y)=-9(\because {(a-b)}^2=a^2+b^2-2ab)$

$\Rightarrow {4+y}^2-4y-{9-y}^2+6y=-9$

$\Rightarrow 2y-5=-9$

$\Rightarrow 2y=-9+5$

$\Rightarrow 2y=-4$

$\Rightarrow y=\frac{-4}{2}=-2$

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