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Q. 43

Expert-verified
Found in: Page 527

### Geometry

Book edition Student Edition
Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Pages 227 pages
ISBN 9780395977279

# 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.

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

See the step by step solution

## 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.

$d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\text{}\left[d=\text{distance}\right]\phantom{\rule{0ex}{0ex}}\text{So.}$

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

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

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

For an isosceles triangle:

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

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

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

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

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

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

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

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

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

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