Chapter 0.1, Problem 30E

(a)

To determine

To write the equation of the line passing through the point $P\left(-2,2\right)$ and parallel to the line $L:2x+y=4$ .

Answer to Problem 30E

The equation of the line parallel to $L:2x+y=4$ and passing through the point $P\left(-2,2\right)$ :

  $y=-2x-2$ .

Explanation of Solution

Given:

Given a point $P\left(-2,2\right)$ and a line $L:2x+y=4$ .

Concept Used:

Slope-Intercept form of a line:

Suppose that the equation of a line is given in the slope-intercept form: $y=mx+c$ .

Then, the $m$ gives the slope of the line and $c$ gives the $y$ -intercept of the line.

Point-Slope equation of a line:

Given the slope $m$ of a line and a point $P\left(x_1,y_1\right)$ through which the line passes, the Point-Slope equation of the line is as follows:

  $y=y_1+m\left(x-x_1\right)$ .

Calculation:

To begin with, write the equation of the line $L:2x+y=4$ in the slope intercept form:

  $\begin{array}{c}2x+y=4\\ y=4-2x\\ y=-2x+4\end{array}$ .

Compare $y=-2x+4$ with the general slope-intercept form and observe that the line has the slope $m=-2$ .

Now, parallel lines have the same slope.

Thus, the line parallel to $L:2x+y=4$ and passing through the point $P\left(-2,2\right)$ will also have the slope $m=-2$ .

Now, to find the equation of the line passing through the point $P\left(-2,2\right)$ with slope $m=-2$ , substitute $m=-2$ and $\left(x_1,y_1\right)=\left(-2,2\right)$ in the point-slope equation $y=y_1+m\left(x-x_1\right)$ of a line:

  $y=2+\left(-2\right)\left(x-\left(-2\right)\right)$ .

Simplify the equation $y=2+\left(-2\right)\left(x-\left(-2\right)\right)$ :

  $\begin{array}{l}y=2+\left(-2\right)\left(x-\left(-2\right)\right)\\ y=2-2\left(x+2\right)\\ y=2-2x-4\\ y=-2x-2\end{array}$ .

Thus, the equation of the line parallel to $L:2x+y=4$ and passing through the point $P\left(-2,2\right)$ :

  $y=-2x-2$ .

Conclusion:

The equation of the line parallel to $L:2x+y=4$ and passing through the point $P\left(-2,2\right)$ :

  $y=-2x-2$ .

(b)

To determine

To write the equation of the line passing through the point $P\left(-2,2\right)$ and perpendicular to the line $L:2x+y=4$ .

Answer to Problem 30E

The equation of the line perpendicular to $L:2x+y=4$ and passing through the point $P\left(-2,2\right)$ :

  $y=\frac{1}{2}x+3$ .

Explanation of Solution

Given:

Given a point $P\left(-2,2\right)$ and a line $L:2x+y=4$ .

Concept Used:

Slope-Intercept form of a line:

Suppose that the equation of a line is given in the slope-intercept form: $y=mx+c$ .

Then, the $m$ gives the slope of the line and $c$ gives the $y$ -intercept of the line.

Slopes of perpendicular lines:

Suppose that two lines are perpendicular and that they have the slopes $m$ and $n$ . Then, it follows that, $mn=-1$ .

Point-Slope equation of a line:

Given the slope $n$ of a line and a point $P\left(x_1,y_1\right)$ through which the line passes, the Point-Slope equation of the line is as follows:

  $y=y_1+n\left(x-x_1\right)$ .

Calculation:

To begin with, write the equation of the line $L:2x+y=4$ is slope intercept form:

  $\begin{array}{c}2x+y=4\\ y=4-2x\\ y=-2x+4\end{array}$ .

Compare $y=-2x+4$ with the general slope-intercept form and observe that the line has the slope $m=-2$ .

Now, let the line perpendicular to $L:2x+y=4$ and passing through the point $P\left(-2,2\right)$ has the slope $n$ .

Then, it follows that:

  $mn=-1$ .

Substitute $m=-2$ and find the value of $n$ :

  $\begin{array}{c}\left(-2\right)n=-1\\ n=\frac{-1}{-2}\\ n=\frac{1}{2}\end{array}$ .

Now, to find the equation of the line passing through the point $P\left(-2,2\right)$ with slope $n=\frac{1}{2}$ , substitute $n=\frac{1}{2}$ and $\left(x_1,y_1\right)=\left(-2,2\right)$ in the point-slope equation $y=y_1+n\left(x-x_1\right)$ of a line:

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

Thus, the equation of the line perpendicular to $L:2x+y=4$ and passing through the point $P\left(-2,2\right)$ :

  $y=\frac{1}{2}x+3$ .

Conclusion:

The equation of the line perpendicular to $L:2x+y=4$ and passing through the point $P\left(-2,2\right)$ :

  $y=\frac{1}{2}x+3$ .