Find the point P(x, y) if its distance from (-3, 0) & (3, 0) is 4 units individually

Distance

between two points $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated using the formula $\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } }$

Given,

Distance

between the points $(x,y) ; (-3,0) =$ 4

between $(x,y) ; (3,0)= 4$

Solving,we get

$=> 12x = 0$

$=> x = 0$

Now,

Distance

between the points $(x,y) ; (-3,0) = 4$

$=> \sqrt { \left( -3-x \right) ^{ 2 }+\left( 0 - y \right) ^{ 2 } } = 4$

But $x = 0$

$=> \sqrt { 9 + { y }^{ 2 } } = 4$

$=> 9 + {y}^{2} = 16$

${y}^{2} = 7$

$=> y = \sqrt {7}$ or $- \sqrt {7}$