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

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

• A

  $$P(1,3)$$
  

• B

  $$P(0,2)$$

• C

  $$P(1,2)$$

• D

  $$P(0,3)$$

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

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

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

$$ \sqrt { { y}^{ 2 } - 2y + 10 } = \sqrt { { y }^{ 2 } - 10y + 26 } $$$$ => { y }^{ 2 } - 2y + 10 = { y }^{ 2 } - 10y + 26  $$$$ 8y = 16 $$$$ y = 2 $$Thus, the point is $$ (0,2) $$