What point on y-axis is equidistant from the points $(3,1)$ and $(1,5)$?

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

Distance Formula

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

Distance between $(0,y)$ and $(3,1) = \sqrt { \left( 3-0 \right) ^{ 2 }+\left( 1 - y \right) ^{ 2 } } = \sqrt { 9 + { 1}^{ 2 }+ { y }^{ 2 } - 2y } = \sqrt { { y }^{ 2 } - 2y + 10 }$

Distance between $(0,y)$ and $(1,5) = \sqrt { \left( 1-0 \right) ^{ 2 }+\left( 5 - y \right) ^{ 2 } } = \sqrt { 1 + { 5}^{ 2 }+ { y }^{ 2 } - 10y } = \sqrt { { y }^{ 2 } - 10y + 26 }$

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

between $(0,y) ; (3,1)$ and $(0,y) ; (1,5)$ are equal.

$\sqrt { { y}^{ 2 } - 2y + 10 } = \sqrt { { y }^{ 2 } - 10y + 26 }$

$\Rightarrow { y }^{ 2 } - 2y + 10 = { y }^{ 2 } - 10y + 26$

$8y = 16$

$y = 2$

Thus, the point is $(0,2)$