Question

Which point on y-axis is equidistant from $(2,3)$ and $(-4,1)$?

Answer 1

Let the point on the y axis be (0,k)

Given that,

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

$\therefore$ $\sqrt{{(2-0)}^2+{(3-k)}^2}=\sqrt{{(-4-0)}^2+{(1-k)}^2}$

Squaring both sides

$4+9+k^2+6k=16+1+k^2-2k$

$8k=17-13$

$8k=4$

$k=\frac{1}{2}$

So point = $(0,\frac{1}{2})$