Q. 43 - Geometry

Question

It is known that triangle GHM is isosceles. G is point (-2, -3), H is point (-2, 7), and the x-coordinate of M is 4. Find all five possible values for the y-coordinate of M.

Step-by-Step Solution

Answer

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

Step-1 – Given

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

Step-2 – To determine

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

Step-3 – Calculation

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:

$\begin{array}{l}{\left(10\right)}^2={\left(\sqrt{y^2+6y+45}\right)}^2\\ 100=y^2+6y+45\\ y^2+6y+45-100=0\\ y^2+6y-55=0\\ y^2+\left(11y-5y\right)-55=0\left[\text{write, 6}y=\left(11y-5y\right)\right]\\ y^2+11y-5y-55=0\\ y\left(y+11\right)-5\left(y+11\right)=0\left[\begin{array}{l}\text{the first and second term has a common factor,}y\text{ and}\\ \text{the third and fourth term has a common factor,5}\end{array}\right]\\ \left(y+11\right)\left(y-5\right)=0\\ \left(y+11\right)=0\text{ or }\left(y-5\right)=0\\ y=-11\text{ or }y=5\end{array}$

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

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

$\begin{array}{l}{\left(10\right)}^2={\left(\sqrt{y^2-14y+85}\right)}^2\\ 100=y^2-14y+85\\ y^2-14y+85-100=0\\ y^2-14y-15=0\\ y^2-\left(15y-y\right)-15=0\left[\text{write, 14}y=\left(15y-y\right)\right]\\ y^2-15y+y-15=0\\ y\left(y-15\right)+1\left(y-15\right)=0\left[\begin{array}{l}\text{the first and second term has a common factor,}y\text{ and}\\ \text{the third and fourth term has a common factor,1}\end{array}\right]\\ \left(y-15\right)\left(y+1\right)=0\\ \left(y-15\right)=0\text{ or }\left(y+1\right)=0\\ y=15\text{ or }y=-1\end{array}$

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

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

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

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