Find a relation between x and y such that the point $(x, y)$ is equidistant from the point $(3, 6)$ and $(-3, 4)$.

Distance between two points $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated

using the formula $\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } }$

Distance between $P (x,y)$ and $A (3,6) = PA = \sqrt { \left( 3-x \right) ^{ 2 }+\left( 6 - y \right) ^{ 2 } } = \sqrt { { 3}^{ 2 }+ { x }^{ 2 } - 6x + { 6}^{ 2 }+ { y }^{ 2 } - 12y } = \sqrt { { x }^{ 2 } - 6x + { y }^{ 2 } - 12y + 45 }$

Distance between $P (x,y)$ and $B (-3,4) = PB = \sqrt { \left( -3-x \right) ^{ 2 }+\left( 4 - y \right) ^{ 2 } } = \sqrt { 9 + { x }^{ 2 } + 6x + 16 + { y }^{ 2 } - 8y } = \sqrt { { x }^{ 2 } + 6x + { y }^{ 2 } - 8y + 25 }$

Given, PA $=$ PB $\sqrt { { x }^{ 2 } - 6x + { y }^{ 2 } - 12y + 45 } = \sqrt { { x }^{ 2 } + 6x + { y }^{ 2 } - 8y + 25 }$

$=> { x }^{ 2 } - 6x + { y }^{ 2 } - 12y + 45 = { x }^{ 2 } + 6x + { y }^{ 2 } - 8y + 25$

$12x + 4y - 20 = 0$

$3x + y - 5 = 0$