Chapter 4.3, Problem 23E

(a)

To determine

To find: an equation of the line that passes through the given point and is parallel to given line.

Answer to Problem 23E

Therefore, the equation of the line that passes through point $\left(4,3\right)$ and parallel to give line $y=-4x+19$

Explanation of Solution

Given:

Consider the graph and the point $\left(4,3\right)$ shown below.

  

Calculation:

Use points $\left(1,6\right)$ and $\left(2,2\right)$ and find the slope $m$ for the given line.

  $\begin{array}{l}m=\frac{2-6}{2-1}\\ \text{ =}-4\end{array}$

Slope of the line is $m=-4$

Find the line parallel to give line

So, the slope of the line will be $m=-4$

Now, use the slope and point $\left(4,3\right)$ to find the equation of line

  $y-y_1=m\left(x-x_1\right)$

Substitute the values

  $\begin{array}{l}y-3=-4\left(x-4\right)\\ y-3=-4x+16\end{array}$

Add 3 on both sides

  $y=-4x+19$

Conclusion:

Therefore, the equation of the line that passes through point $\left(4,3\right)$ and parallel to give line $y=-4x+19$

(b)

To determine

To find: an equation of the line that passes through the given point and is perpendicular to given line.

Answer to Problem 23E

Therefore, the equation of the line that passes through point $\left(4,3\right)$ and perpendicular to give line is $y=\frac{1}{4}x+2$

Explanation of Solution

Given:

Consider the graph and the point $\left(4,3\right)$ shown below.

  

Calculation:

Use points $\left(1,6\right)$ and $\left(2,2\right)$ and find the slope $m$ for the given line.

  $\begin{array}{l}m=\frac{2-6}{2-1}\\ \text{ =}-4\end{array}$

Slope of the line is $m=-4$

Find the line perpendicular to given the

So, the slope of the line will be $m=\frac{1}{4}$

Now, use the slope and point $\left(4,3\right)$ to find the equation of the

  $y-y_1=m\left(x-x_1\right)$

Substitute the values

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

Add 3 on both sides, so

  $y=\frac{1}{4}x+2$

Therefore, the equation of the line that passes through point $\left(4,3\right)$ and parallel to given line is $y=-4x+19$ and perpendicular to give line is $y=\frac{1}{4}x+2$

Conclusion:

Therefore, the equation of the line that passes through point $\left(4,3\right)$ and perpendicular to give line is $y=\frac{1}{4}x+2$