91Ó°ÊÓ

The given points are$G\left(−2,−3\right)\text{,}H\left(−2,7\right)$

We have to find all possible values for the y-coordinate of M.

Let us assume, the y-coordinate of M = y.

So, the coordinate of M is of the form $M\left(4,y\right)$.

We’ll use the distance formula to find the lengths GH, GM and HM.

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

$\begin{array}{l}GH=\sqrt{{\left(−2−\left(−2\right)\right)}^2+{\left(7−\left(−3\right)\right)}^2}\left[\text{since,}G\left(−2,−3\right)\text{and}H\left(−2,7\right)\right]\\ GH=\sqrt{{\left(−2+2\right)}^2+{\left(7+3\right)}^2}\\ GH=\sqrt{{\left(0\right)}^2+{\left(10\right)}^2}\\ GH=\sqrt{0+100}\\ GH=\sqrt{100}\\ GH=10\end{array}$

$\begin{array}{l}GM=\sqrt{{\left(4−\left(−2\right)\right)}^2+{\left(y−\left(−3\right)\right)}^2}\left[\text{since,}G\left(−2,−3\right)\text{and}M\left(4,y\right)\right]\\ GM=\sqrt{{\left(4+2\right)}^2+{\left(y+3\right)}^2}\\ GM=\sqrt{{\left(6\right)}^2+{\left(y+3\right)}^2}\\ GM=\sqrt{36+y^2+2\left(y\right)\left(3\right)+{\left(3\right)}^2}\\ GM=\sqrt{36+y^2+6y+9}\\ GM=\sqrt{y^2+6y+45}\end{array}$

$\begin{array}{l}HM=\sqrt{{\left(4−\left(−2\right)\right)}^2+{\left(y−7\right)}^2}\left[\text{since,}H\left(−2,7\right)\text{and}M\left(4,y\right)\right]\\ HM=\sqrt{{\left(4+2\right)}^2+{\left(y−7\right)}^2}\\ HM=\sqrt{{\left(6\right)}^2+{\left(y−7\right)}^2}\\ HM=\sqrt{36+y^2−2\left(y\right)\left(7\right)+{\left(7\right)}^2}\\ HM=\sqrt{36+y^2−14y+49}\\ HM=\sqrt{y^2−14y+85}\end{array}$

For an isosceles triangle:

${\left(GH\right)}^2={\left(GM\right)}^2$

Plug the values of GH and GM in the above equation:

$\begin{array}{l}{\left(10\right)}^2={\left(\sqrt{y^2+6y+45}\right)}^2\\ 100=y^2+6y+45\\ y^2+6y+45−100=0\\ y^2+6y−55=0\\ y^2+\left(11y−5y\right)−55=0\left[\text{write,6}y=\left(11y−5y\right)\right]\\ y^2+11y−5y−55=0\\ y\left(y+11\right)−5\left(y+11\right)=0\left[\begin{array}{l}\text{thefirstandsecondtermhasacommonfactor,}y\text{and}\\ \text{thethirdandfourthtermhasacommonfactor,5}\end{array}\right]\\ \left(y+11\right)\left(y−5\right)=0\\ \left(y+11\right)=0\text{or}\left(y−5\right)=0\\ y=−11\text{or}y=5\end{array}$

${\left(GH\right)}^2={\left(HM\right)}^2$

Plug the values of GH and HM in the above equation:

$\begin{array}{l}{\left(10\right)}^2={\left(\sqrt{y^2−14y+85}\right)}^2\\ 100=y^2−14y+85\\ y^2−14y+85−100=0\\ y^2−14y−15=0\\ y^2−\left(15y−y\right)−15=0\left[\text{write,14}y=\left(15y−y\right)\right]\\ y^2−15y+y−15=0\\ y\left(y−15\right)+1\left(y−15\right)=0\left[\begin{array}{l}\text{thefirstandsecondtermhasacommonfactor,}y\text{and}\\ \text{thethirdandfourthtermhasacommonfactor,1}\end{array}\right]\\ \left(y−15\right)\left(y+1\right)=0\\ \left(y−15\right)=0\text{or}\left(y+1\right)=0\\ y=15\text{or}y=−1\end{array}$

${\left(GM\right)}^2={\left(HM\right)}^2$

Plug the values of GM and HM in the above equation:

$\begin{array}{l}{\left(\sqrt{y^2+6y+45}\right)}^2={\left(\sqrt{y^2−14y+85}\right)}^2\\ y^2+6y+45=y^2−14y+85\\ 6y+45+14y−85=0\\ 20y−40=0\\ 20y=40\\ y=2\left[\text{dividebothsideby20}\right]\end{array}$

So, the five possible values for the y-coordinate of M are $−11,5,15,−1,2$.