Chapter 15, Problem 25CR

a.

To determine

To find: the length of the median to $\overline{AB}$ .

Answer to Problem 25CR

  $\sqrt{26}$ .

Explanation of Solution

Given:

The vertices of $\Delta ABC$ are $A=\left(5,4\right),B=\left(11,6\right)\text{ and }C=\left(9,10\right)$ .

Calculation:

According to the question, the vertices of $\Delta ABC$ are $A=\left(5,4\right),B=\left(11,6\right)\text{ and }C=\left(9,10\right)$ .

The figure of this condition is:

  

Now,

  $\begin{array}{l}\text{mid-point }\overline{AB}=\left(\frac{5+11}{2},\frac{4+6}{2}\right)\\ =\left(8,5\right)\end{array}$

Thus, the length of the median to $\overline{AB}$ is

  $\begin{array}{l}\text{length}=\sqrt{{\left(9-8\right)}^2+{\left(10-5\right)}^2}\\ =\sqrt{1+25}=\sqrt{26}\end{array}$

Hence, the length of the median to $\overline{AB}$ is $\sqrt{26}$ .

b.

To determine

To find: the equation of the median to $\overline{AB}$ .

Answer to Problem 25CR

  $y=5x-35$ .

Explanation of Solution

Given:

The vertices of $\Delta ABC$ are $A=\left(5,4\right),B=\left(11,6\right)\text{ and }C=\left(9,10\right)$ .

Calculation:

According to the question, the vertices of $\Delta ABC$ are $A=\left(5,4\right),B=\left(11,6\right)\text{ and }C=\left(9,10\right)$ .

The figure of this condition is:

  

Now, $\text{slope median }\overline{AB}=\frac{10-5}{9-8}=5$

Then, the equation of the median to $\overline{AB}$ is:

  $\begin{array}{l}y=mx+b\left\{\text{equation of the line}\right\}\\ 5=5\left(8\right)+b\left\{\text{substitutes these value}\right\}\\ 5=40+b\\ \therefore b=5-40=-35\\ \therefore y=5x-35\end{array}$

Hence, the equation of the median to $\overline{AB}$ is $y=5x-35$ .

c.

To determine

To find: the equation of the altitude to $\overline{AB}$ .

Answer to Problem 25CR

  $y=-3x+37$ .

Explanation of Solution

Given:

The vertices of $\Delta ABC$ are $A=\left(5,4\right),B=\left(11,6\right)\text{ and }C=\left(9,10\right)$ .

Calculation:

According to the question, the vertices of $\Delta ABC$ are $A=\left(5,4\right),B=\left(11,6\right)\text{ and }C=\left(9,10\right)$ .

The figure of this condition is:

  

Now, $\text{slope }\overline{AB}=\frac{4-6}{5-11}=\frac{-2}{-6}=\frac{1}{3}$

So, slope altitude to $\overline{AB}$ is $\overline{AB}=-3\{m_1{m}_2=-1\}$ .

Then, the equation of the altitude to $\overline{AB}$ is:

  $\begin{array}{l}y=mx+b\left\{\text{equation of the line}\right\}\\ 10=\left(-3\right)\left(9\right)+b\left\{\text{substitutes these value}\right\}\\ 10=-27+b\\ \therefore b=10+27=37\\ \therefore y=-3x+37\end{array}$

Hence, the equation of the altitude to $\overline{AB}$ is $y=-3x+37$ .

d.

To determine

To find: the equation of the perpendicular bisector of $\overline{AB}$ .

Answer to Problem 25CR

  $y=-3x+29$ .

Explanation of Solution

Given:

The vertices of $\Delta ABC$ are $A=\left(5,4\right),B=\left(11,6\right)\text{ and }C=\left(9,10\right)$ .

Calculation:

According to the question, the vertices of $\Delta ABC$ are $A=\left(5,4\right),B=\left(11,6\right)\text{ and }C=\left(9,10\right)$ .

The figure of this condition is:

  

Now, slope of perpendicular bisector of $\overline{AB}$ is $\overline{AB}=-3$ .

  $\text{mid-point }\overline{AB}=\left(8,5\right)\text{{from}\text{part a}\}$

Now, the equation of the perpendicular bisector to $\overline{AB}$ is:

  $\begin{array}{l}y=mx+b\left\{\text{equation of the line}\right\}\\ 5=\left(-3\right)\left(8\right)+b\left\{\text{substitutes these value}\right\}\\ 5=-24+b\\ \therefore b=5+24=29\\ \therefore y=-3x+29\end{array}$

Hence, the equation of the median to $\overline{AB}$ is $y=-3x+29$ .