Chapter 0.1, Problem 29E

(a)

To determine

To write the equation of the line passing through the point $P\left(0,0\right)$ and parallel to the line $L:y=-x+2$ .

Answer to Problem 29E

The equation of the line parallel to $L:y=-x+2$ and passing through the point $P\left(0,0\right)$ :

  $y=-x$ .

Explanation of Solution

Given:

Given a point $P\left(0,0\right)$ and a line $L:y=-x+2$ .

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:

Observe that the line $L:y=-x+2$ is already in the slope-intercept form.

Compare $y=-x+2$ with the general slope-intercept form and observe that the line has the slope $m=-1$ .

Now, parallel lines have the same slope.

Thus, the line parallel to $L:y=-x+2$ and passing through the point $P\left(0,0\right)$ will also have the slope $m=-1$ .

Now, to find the equation of the line passing through the point $P\left(0,0\right)$ with slope $m=-1$ , substitute $m=-1$ and $\left(x_1,y_1\right)=\left(0,0\right)$ in the point-slope equation $y=y_1+m\left(x-x_1\right)$ of a line:

  $y=0+\left(-1\right)\left(x-0\right)$ .

Simplify the equation $y=0+\left(-1\right)\left(x-0\right)$ :

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

Thus, the equation of the line parallel to $L:y=-x+2$ and passing through the point $P\left(0,0\right)$ :

  $y=-x$ .

Conclusion:

The equation of the line parallel to $L:y=-x+2$ and passing through the point $P\left(0,0\right)$ :

  $y=-x$ .

To determine

To write the equation of the line passing through the point $P\left(0,0\right)$ and perpendicular to the line $L:y=-x+2$ .

Answer to Problem 29E

The equation of the line perpendicular to $L:y=-x+2$ and passing through the point $P\left(0,0\right)$ :

  $y=x$ .

Explanation of Solution

Given:

Given a point $P\left(0,0\right)$ and a line $L:y=-x+2$ .

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:

Observe that the line $L:y=-x+2$ is already in the slope-intercept form.

Compare $y=-x+2$ with the general slope-intercept form and observe that the line has the slope $m=-1$ .

Now, let the line perpendicular to $L:y=-x+2$ and passing through the point $P\left(0,0\right)$ has the slope $n$ .

Then, it follows that:

  $mn=-1$ .

Substitute $m=-1$ and find the value of $n$ :

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

Now, to find the equation of the line passing through the point $P\left(0,0\right)$ with slope $n=1$ , substitute $n=1$ and $\left(x_1,y_1\right)=\left(0,0\right)$ in the equation $y=y_1+n\left(x-x_1\right)$ :

  $\begin{array}{c}y=0+1\left(x-0\right)\\ y=x\end{array}$ .

Thus, the equation of the line perpendicular to $L:y=-x+2$ and passing through the point $P\left(0,0\right)$ :

  $y=x$ .

Conclusion:

The equation of the line perpendicular to $L:y=-x+2$ and passing through the point $P\left(0,0\right)$ :

  $y=x$ .