Chapter 15, Problem 11CR

a.

To determine

To find the lengths and the slopes of AB, BC and AC.

Answer to Problem 11CR

  $13,14,15$

Explanation of Solution

Given:

The given triangle ABC with vertices $A(3,8),B(8,-4)$ and $C(-6,-4)$ .

Calculation:

Calculating the lengths of AB, BC and AC as follows-

  $\begin{array}{l}AB=\sqrt{{(3-8)}^2+{(8+4)}^2}=\sqrt{25+144}=13\\ BC=\sqrt{{(8+6)}^2+{(-4+4)}^2}=\sqrt{{14}^2}=14\\ AC=\sqrt{{(3+6)}^2+{(8+4)}^2}=\sqrt{81+144}=15\end{array}$

b.

To determine

To find the type of triangle.

Answer to Problem 11CR

Acute.

Explanation of Solution

Given:

The given triangle ABC with vertices $A(3,8),B(8,-4)$ and $C(-6,-4)$ .

Calculation:

Since from part (a) all the sides are different, the triangle is acute.

c.

To determine

To find the equation of AC and its x and y intercepts.

Answer to Problem 11CR

  $y=\frac{4}{3}x+4$

Explanation of Solution

Given:

The given triangle ABC with vertices $A(3,8),B(8,-4)$ and $C(-6,-4)$ .

Calculation:

The equation of AC is

  $\begin{array}{l}y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\ y-8=\frac{-4-8}{-6-3}(x-3)\\ y-8=\frac{4}{3}(x-3)\\ y=\frac{4}{3}x+4\end{array}$

d.

To determine

To find the equation of BC.

Answer to Problem 11CR

  $y=-4$

Explanation of Solution

Given:

The given triangle ABC with vertices $A(3,8),B(8,-4)$ and $C(-6,-4)$ .

Calculation:

The equation of BC is

  $\begin{array}{l}y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\ y+4=\frac{-4+4}{-6-8}(x-8)\\ y+4=0\\ y=-4\end{array}$

e.

To determine

To find where does the altitude BC intersect BC.

Explanation of Solution

Given:

The given triangle ABC with vertices $A(3,8),B(8,-4)$ and $C(-6,-4)$ .

Calculation:

Calculating the lengths of AB, BC and AC as follows-

  $\begin{array}{l}AB=\sqrt{{(3-8)}^2+{(8+4)}^2}=\sqrt{25+144}=13\\ BC=\sqrt{{(8+6)}^2+{(-4+4)}^2}=\sqrt{{14}^2}=14\\ AC=\sqrt{{(3+6)}^2+{(8+4)}^2}=\sqrt{81+144}=15\end{array}$

f.

To determine

To find the equation of BC.

  $y=-4$

Given:

The given triangle ABC with vertices $A(3,8),B(8,-4)$ and $C(-6,-4)$ .

Calculation:

The equation of BC is

  $\begin{array}{l}y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\ y+4=\frac{-4+4}{-6-8}(x-8)\\ y+4=0\\ y=-4\end{array}$

Answer to Problem 11CR

  $y=-4$

Explanation of Solution

Given:

The given triangle ABC with vertices $A(3,8),B(8,-4)$ and $C(-6,-4)$ .

Calculation:

The equation of BC is

  $\begin{array}{l}y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\ y+4=\frac{-4+4}{-6-8}(x-8)\\ y+4=0\\ y=-4\end{array}$

g.

To determine

To find the lengths of altitude to BC.

Answer to Problem 11CR

  $12$

Explanation of Solution

Given:

The given triangle ABC with vertices $A(3,8),B(8,-4)$ and $C(-6,-4)$ .

Calculation:

The length of altitude to BC is $12$ .

h.

To determine

To find the midpoint of CB and the slope of the median to CB.

Answer to Problem 11CR

  $(1,-4)$

Explanation of Solution

Given:

The given triangle ABC with vertices $A(3,8),B(8,-4)$ and $C(-6,-4)$ .

Calculation:

The midpoint of CB is

  $\left(\frac{8-6}{2},\frac{-4-4}{2}\right)=(1,-4)$

i.

To determine

To find the area of triangle ABC,

Answer to Problem 11CR

  $84$

Explanation of Solution

Given:

The given triangle ABC with vertices $A(3,8),B(8,-4)$ and $C(-6,-4)$ .

Calculation:

The area of triangle ABC is-

  $\frac{ab\sin c}{2}=84$