Chapter 13.2, Problem 8PSA

(a)

To determine

To find: The equation of line containing $\left(2,1\right)$ and $\left(3,4\right)$ .

Answer to Problem 8PSA

The equation of line is $y=3x-5$ .

Explanation of Solution

Given:

The points are $\left(2,1\right)$ and $\left(3,4\right)$ .

Calculation:

Find the equation of line as follows.

Substitute $2$ for $x_1$ , 3 for $x_2$ , $4$ for $y_2$ and $1$ for $y_1$ in the equation $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$ .

  $\begin{array}{c}y-1=\frac{4-1}{3-2}\left(x-2\right)\\ y-1=3\left(x-2\right)\\ y-1=3x-6\\ y=3x-5\end{array}$

Therefore, the equation of line is $y=3x-5$ .

(b)

To determine

To find: The equation of line containing $\left(-6,3\right)$ and $\left(2,-1\right)$ .

Answer to Problem 8PSA

The equation of line is $y=-\frac{1}{2}x$ .

Explanation of Solution

Given:

The points are $\left(-6,3\right)$ and $\left(2,-1\right)$ .

Calculation:

Find the equation of line as follows.

Substitute $-6$ for $x_1$ , 2 for $x_2$ , $-1$ for $y_2$ and $3$ for $y_1$ in the equation $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$ .

  $\begin{array}{c}y-3=\frac{-1-3}{2+6}\left(x+6\right)\\ y-3=-\frac{1}{2}\left(x+6\right)\\ y-3=-\frac{1}{2}x-3\\ y=-\frac{1}{2}x\end{array}$

Therefore, the equation of line is $y=-\frac{1}{2}x$ .

(c)

To determine

To find: The equation of line containing $\left(1,5\right)$ and $\left(-3,5\right)$ .

Answer to Problem 8PSA

The equation of line is $y=5$ .

Explanation of Solution

Given:

The points are $\left(1,5\right)$ and $\left(-3,5\right)$ .

Calculation:

Find the equation of line as follows.

Substitute $1$ for $x_1$ , $-3$ for $x_2$ , $5$ for $y_2$ and $5$ for $y_1$ in the equation $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$ .

  $\begin{array}{c}y-5=\frac{5-5}{-3-1}\left(x-1\right)\\ y-5=0\\ y=5\end{array}$

Therefore, the equation of line is $y=5$ .

(d)

To determine

To find: The equation of line with an x-intercepts of 2 and slope $7$ .

Answer to Problem 8PSA

The equation of line is $y=7x-14$ .

Explanation of Solution

Given:

The line with an x-intercepts of 2 and slope $7$ .

Calculation:

The x-intercept is 2 that means the point is $\left(2,0\right)$ .

Find the equation of line as follows.

Substitute $2$ for $x_1$ , 0 for $y_1$ and 7 for $m$ in the equation $y-y_1=m\left(x-x_1\right)$ .

  $\begin{array}{c}y-0=7\left(x-2\right)\\ y=7x-14\end{array}$

Therefore, the equation of line is $y=7x-14$ .

(e)

To determine

To find: The equation of line that has an x-intercept of $3$ and passes through $\left(1,8\right)$ .

Answer to Problem 8PSA

The equation of line is $y=-4\left(x-3\right)$ .

Explanation of Solution

Given:

The line has an x-intercept of $3$ and passes through $\left(1,8\right)$ .

Calculation:

Find the equation of line as follows.

Substitute $3$ for $x_1$ , $1$ for $x_2$ , $8$ for $y_2$ and $0$ for $y_1$ in the equation $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$ .

  $\begin{array}{c}y-0=\frac{8-0}{1-3}\left(x-3\right)\\ y=-4\left(x-3\right)\end{array}$

Therefore, the equation of line is $y=-4\left(x-3\right)$ .

(f)

To determine

To find: The equation of line that passes through $\left(-3,6\right)$ and $\left(-3,10\right)$ .

Answer to Problem 8PSA

This is a vertical line $x=-3$ .

Explanation of Solution

Given:

The points are $\left(-3,6\right)$ and $\left(-3,10\right)$ .

Calculation:

Find the equation of line as follows.

Substitute $-3$ for $x_1$ , $-3$ for $x_2$ , $10$ for $y_2$ and $6$ for $y_1$ in the equation $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$ .

  $\begin{array}{c}y-6=\frac{10-6}{-3+3}\left(x-1\right)\\ y-6=\text{undefined}\end{array}$

Therefore, this is a vertical line $x=-3$ .

(g)

To determine

To find: The equation of line that passes through $\left(8,7\right)$ and is perpendicular to the graph of $3y=-2x+24$ .

Answer to Problem 8PSA

The equation of line is $y-7=\frac{3}{2}\left(x-8\right)$ .

Explanation of Solution

Given:

The line passes through $\left(8,7\right)$ and is perpendicular to the graph of $3y=-2x+24$ .

Calculation:

Rewrite the equation $3y=-2x+24$ as follows.

  $\begin{array}{c}3y=-2x+24\\ y=-\frac{2}{3}x+8\end{array}$

Compare the equation $y=-\frac{2}{3}x+8$ with $y=mx+b$ .

  $\begin{array}{c}m=-\frac{2}{3}\\ b=8\end{array}$

The line is perpendicular so $m_2=-\frac{1}{m_1}$ .

  $m_2=\frac{3}{2}$

Substitute 8 for $x_1$ , 7 for $y_1$ and $\frac{3}{2}$ for $m$ in the equation $y-y_1=m\left(x-x_1\right)$ .

  $y-7=\frac{3}{2}\left(x-8\right)$

Therefore, the equation of line is $y-7=\frac{3}{2}\left(x-8\right)$ .