Chapter 13.2, Problem 6PSA

(a)

To determine

To write: The y-form equation of the line with y-intercept of 2 and slope 4.

Answer to Problem 6PSA

The y-form equation is $y=4x+2$ .

Explanation of Solution

Given:

The y-intercept is $2$ and slope is $4$ .

Calculation:

Substitute $4$ for $m$ , and $2$ for $b$ in the equation $y=mx+b$ .

  $y=4x+2$

Therefore, the y-form equation is $y=4x+2$ .

(b)

To determine

To write: The y-form equation of the line with slope 5 and passes through $\left(0,-2\right)$ .

Answer to Problem 6PSA

The y-form equation is $y=5x-2$ .

Explanation of Solution

Given:

The slope is $4$ and point is $\left(0,-2\right)$ .

Calculation:

Substitute $4$ for $m$ , 0 for $x_1$ , and $-2$ for $y_1$ in the equation $y-y_1=m\left(x-x_1\right)$ .

  $\begin{array}{c}y+2=5\left(x-0\right)\\ y+2=5x\\ y=5x-2\end{array}$

Therefore, the y-form equation is $y=5x-2$ .

(c)

To determine

To write: The y-form equation of the line parallel to graph of $y=10x-6$ and y-intercept of 1.

Answer to Problem 6PSA

The y-form equation is $y=10x+1$ .

Explanation of Solution

Given:

It is given that line parallel to graph of $y=10x-6$ and y-intercept of 1.

Calculation:

Compare the equation $y=10x-6$ with $y=mx+b$ .

  $m_1=10$

As the line parallel so $m_1=m_2$ .

  $m_2=10$

Substitute 10 for $m$ , and 1 for $b$ in the equation $y=mx+b$ .

  $y=10x+1$

Therefore, the y-form equation is $y=10x+1$ .

(d)

To determine

To write: The y-form equation of the line perpendicular to graph of $2y=x+16$ and passes through $\left(0,-5\right)$ .

Answer to Problem 6PSA

The y-form equation is $y=-2x-5$ .

Explanation of Solution

Given:

It is given that line perpendicular to graph of $2y=x+16$ and passes through $\left(0,-5\right)$ .

Calculation:

Rewrite the equation $2y=x+16$ as follows.

  $\begin{array}{c}2y=x+16\\ y=\frac{x}{2}+8\end{array}$

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

  $m_1=\frac{1}{2}$

As the line perpendicular so $m_2=-\frac{1}{m_1}$ .

  $m_2=-2$

Substitute $-2$ for $m$ , 0 for $x_1$ , and $-5$ for $y_1$ in the equation $y-y_1=m\left(x-x_1\right)$ .

  $\begin{array}{c}y+5=-2\left(x-0\right)\\ y+5=-2x\\ y=-2x-5\end{array}$

Therefore, the y-form equation is $y=-2x-5$ .

(e)

To determine

To write: The y-form equation of the line with y-intercept 2, and line perpendicular to line containing $\left(-4,6\right)$ and $\left(1,11\right)$ .

Answer to Problem 6PSA

The y-form equation is $y=-x+2$ .

Explanation of Solution

Given:

It is given that line has y-intercept 2, and line perpendicular to line containing $\left(-4,6\right)$ and $\left(1,11\right)$ .

Calculation:

Substitute $-4$ for $x_1$ , 1 for $x_2$ , $11$ for $y_2$ and $6$ for $y_1$ in the equation $m=\frac{y_2-y_1}{x_2-x_1}$ .

  $\begin{array}{c}m=\frac{11-6}{1-\left(-4\right)}\\ =\frac{5}{5}\\ =1\end{array}$

The line perpendicular so slope will be $m=-1$ .

Substitute $-1$ for $m$ , and 2 for $b$ in the equation $y=mx+b$ .

  $\begin{array}{c}y=-1x+2\\ y=-x+2\end{array}$

Therefore, the y-form equation is $y=-x+2$ .