Find the coordinates of the point that is equidistant from (-2, 5), (8, 5), and (6, 7).

The given points are:$\left(-2,5\right)\text{,}\left(8,5\right)\text{and}\left(6,7\right)$

We have to find the coordinates of the point that is equidistant from (-2, 5), (8, 5) and (6, 7).

Let us assume that $A\left(-2,5\right)\text{,}B\left(8,5\right)\text{and}C\left(6,7\right)$.

Let, is the point that is equidistant from the given points (-2, 5), (8, 5) and (6, 7).

It means, $AO=BO=CO$.

We’ll use the distance formula to find the value AO, BO and CO.

$$\begin{gathered} d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\left[d=\text{distance}\right] \\ \text{So,} \end{gathered}$$

$\begin{array}{l}AO=\sqrt{{\left(x-\left(-2\right)\right)}^2+{\left(y-5\right)}^2}\left[\text{since,}A\left(-2,5\right)\text{and}O\left(x,y\right)\right]\\ AO=\sqrt{{\left(x+2\right)}^2+{\left(y-5\right)}^2}\left[\text{perfectsquaretrinomial}\right]\\ AO=\sqrt{\left[{\left(x\right)}^2+2\left(x\right)\left(2\right)+{\left(2\right)}^2\right]+\left[{\left(y\right)}^2-2\left(y\right)\left(5\right)+{\left(5\right)}^2\right]}\\ AO=\sqrt{x^2+4x+4+y^2-10y+25}\\ AO=\sqrt{x^2+y^2+4x-10y+29}\end{array}$

$\begin{array}{l}BO=\sqrt{{\left(x-8\right)}^2+{\left(y-5\right)}^2}\left[\text{since,}B\left(8,5\right)\text{and}O\left(x,y\right)\right]\\ BO=\sqrt{{\left(x-8\right)}^2+{\left(y-5\right)}^2}\left[\text{perfectsquaretrinomial}\right]\\ BO=\sqrt{\left[{\left(x\right)}^2-2\left(x\right)\left(8\right)+{\left(8\right)}^2\right]+\left[{\left(y\right)}^2-2\left(y\right)\left(5\right)+{\left(5\right)}^2\right]}\\ BO=\sqrt{x^2-16x+64+y^2-10y+25}\\ BO=\sqrt{x^2+y^2-16x-10y+89}\end{array}$

$\begin{array}{l}CO=\sqrt{{\left(x-6\right)}^2+{\left(y-7\right)}^2}\left[\text{since,}C\left(6,7\right)\text{and}O\left(x,y\right)\right]\\ CO=\sqrt{{\left(x-6\right)}^2+{\left(y-7\right)}^2}\left[\text{perfectsquaretrinomial}\right]\\ CO=\sqrt{\left[{\left(x\right)}^2-2\left(x\right)\left(6\right)+{\left(6\right)}^2\right]+\left[{\left(y\right)}^2-2\left(y\right)\left(7\right)+{\left(7\right)}^2\right]}\\ CO=\sqrt{x^2-12x+36+y^2-14y+49}\\ CO=\sqrt{x^2+y^2-12x-14y+85}\end{array}$

Then,

${\left(AO\right)}^2={\left(BO\right)}^2$

Plug the values of AO and BO in the above equation:

$\begin{array}{l}{\left(\sqrt{x^2+y^2+4x-10y+29}\right)}^2={\left(\sqrt{x^2+y^2-16x-10y+89}\right)}^2\\ x^2+y^2+4x-10y+29=x^2+y^2-16x-10y+89\\ 4x+16x+29-89=0\left[\text{since,}{x}^2,y^2,10y\text{aresame}\right]\\ 20x-60=0\\ 20x=60\\ x=3\left[\text{dividebothsideby20}\right]\end{array}$

${\left(AO\right)}^2={\left(CO\right)}^2$

Plug the values of AO and CO in the above equation:

$\begin{array}{l}{\left(\sqrt{x^2+y^2+4x-10y+29}\right)}^2={\left(\sqrt{x^2+y^2-12x-14y+85}\right)}^2\\ x^2+y^2+4x-10y+29=x^2+y^2-12x-14y+85\\ 4x+12x-10y+14y+29-85=0\left[\text{since,}{x}^2,y^2\text{aresame}\right]\\ 16x+4y-56=0\\ 16\left(3\right)+4y-56=\left[\text{since,}x=3\right]\\ 48+4y-56=0\\ 4y-8=0\\ 4y=8\\ y=2\left[\text{dividebothsideby4}\right]\end{array}$

So, the coordinates of the point that is equidistant from given point is (3, 2).