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

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

Distance between $(0,y)$ and $(-3,4) = \sqrt { \left( -3-0 \right) ^{ 2 }+\left( 4 - y \right) ^{ 2 } } = \sqrt { 9 + { 4}^{ 2 }+ { y }^{ 2 } - 8y } = \sqrt { { y }^{ 2 } - 8y + 25 }$

Distance between $(0,y)$ and $(2,3) = \sqrt { \left( 2-0 \right) ^{ 2 }+\left( 3 - y \right) ^{ 2 } } = \sqrt { 4 + { 3}^{ 2 }+ { y }^{ 2 } - 6y } = \sqrt { { y }^{ 2 } - 6y + 13 }$

As the point $(0,y)$ is equidistant from the two points, both the distances

calculated are equal.

$\sqrt { { y }^{ 2 } - 8y + 25 } = \sqrt { { y }^{ 2 } - 6y + 13 }$

$=> { y }^{ 2 } - 8y + 25 = { y }^{ 2 } - 6y + 13$

$12 = 2y$

$y = 6$

Thus, the point is $(0,6)$