Chapter 4.3, Problem 12Q

(a)

To determine

To write: of the equation of the line which is parallelto the given line.

Answer to Problem 12Q

Therefore, the equation of the line that passes through point $\left(6,2\right)$ and parallel to given line is $y=x+4$

Explanation of Solution

Given:

  

Formula Used:

  $y-y_1=m\left(x-x_1\right)$

Calculation:

Consider the graph and the point $\left(6,2\right)$ shown below:

  

Use points $\left(0,-4\right)and\left(2,2\right)$ and find the slope $m$ for the given line

  $\begin{array}{l}m=\frac{2-\left(-4\right)}{2-0}\\ =\frac{2}{2}\\ =1\end{array}$

Slope of the line is $m=1$

Find the line parallel to give line

So, the slope of the line will be $m=1$

Now, use the slope and point $\left(6,2\right)$ to find the equation of line

  $y-y_1=m\left(x-x_1\right)$

Substitute the values

  $\begin{array}{l}y-2=1\left(x-6\right)\\ y-2=x-6\end{array}$

Add 2 on both sides

  $y=x-4$

Conclusion:

Therefore, the equation of the line that passes through point $\left(6,2\right)$ and perpendicular to the given line is $y=-x+8$

(b)

To determine

To write: of the equation of the line which is parallel to the given line.

Answer to Problem 12Q

The equation of the line that passes through point $\left(6,2\right)$ and perpendicular to the given line is $y=-x+8$

Explanation of Solution

Calculation:

Find the line perpendicular to given line

So, the slope of the line will be $m=-1$

Now, use the slope and point $\left(6,2\right)$ to find the equation of line

  $y-y_1=m\left(x-x_1\right)$

Substitute the values

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

Add $2$ on both sides

  $y=-x+8$

Therefore, the equation of the line that passes through point $\left(6,2\right)$ and perpendicular to the given line is $y=-x+8$

Conclusion:

Therefore, the equation of the line that passes through point $\left(6,2\right)$ and perpendicular to the given line is $y=-x+8$