Find the values of x, y if the distances of the point

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

It is said that (x, y) is equidistant from both (−3,0) and (3,0).

Let  be the distance between (x, y) and (−3,0).

Let  be the distance between (x, y) and (3,0).

So, using the distance formula for both these pairs of points we have

$d_{1}=\sqrt{(x+3)^{2}+(y-0)^{2}}$

$d_{2}=\sqrt{(x-3)^{2}+(y-0)^{2}}$

Now since both these distances are given to be the same, let us equate both  and .

$d_{1}=d_{2}$

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

Squaring on both sides we have,

$(x+3)^{2}+(y-0)^{2}=(x-3)^{2}+(y-0)^{2}$

$x^{2}+9+6 x+y^{2}=x^{2}+9-6 x+y^{2}$

$12 x=0$

$x=0$

It is also said that the value of both  and  is 4 units.

Substituting the value of ‘x’ in the equation for either  or  we can get the value of ‘y’.

$d_{1}=\sqrt{(x+3)^{2}+(y-0)^{2}}$

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

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

Squaring on both sides of the equation we have,

$16=9+y^{2}$

$y^{2}=7$

$y=\pm \sqrt{7}$

Hence the values of ' $x$ ' and ' $y$ ' are $x=0$

$y=\pm \sqrt{7}$ .