Chapter 0.1, Problem 32E

(a)

To determine

To write the equation of the line passing through the point $P\left(-1,\frac{1}{2}\right)$ and parallel to the line $L:y=3$ .

Answer to Problem 32E

The line parallel to $L:y=3$ and passing through the point $P\left(-1,\frac{1}{2}\right)$ will be $y=\frac{1}{2}$ .

Explanation of Solution

Given:

Given a point $P\left(-1,\frac{1}{2}\right)$ and a line $L:y=3$ .

Calculation:

Observe that the line $L:y=3$ is a horizontal one.

Thus, the lines parallel to $L:y=3$ are also horizontal lines.

Now, the horizontal line passing through the point $P\left(-1,\frac{1}{2}\right)$ is $y=\frac{1}{2}$ .

Thus, the line parallel to $L:y=3$ and passing through the point $P\left(-1,\frac{1}{2}\right)$ will be $y=\frac{1}{2}$ .

Conclusion:

The line parallel to $L:y=3$ and passing through the point $P\left(-1,\frac{1}{2}\right)$ will be $y=\frac{1}{2}$ .

(b)

To determine

To write the equation of the line passing through the point $P\left(-1,\frac{1}{2}\right)$ and perpendicular to the line $L:y=3$ .

Answer to Problem 32E

The equation of the line perpendicular to $L:y=3$ and passing through the point $P\left(-1,\frac{1}{2}\right)$ is $x=-1$ .

Explanation of Solution

Given:

Given a point $P\left(-1,\frac{1}{2}\right)$ and a line $L:y=3$ .

Calculation:

Since the line $L:y=3$ is a horizontal one, lines perpendicular to $L:y=3$ are all vertical ones having the general form $x=a$ .

Now, it is required to find the equation of the line passing through the point $P\left(-1,\frac{1}{2}\right)$ .

Now, the equation of the vertical line passing through the point $P\left(-1,\frac{1}{2}\right)$ is $x=-1$ .

Thus, the equation of the line perpendicular to $L:y=3$ and passing through the point $P\left(-1,\frac{1}{2}\right)$ is $x=-1$ .

Conclusion:

The equation of the line perpendicular to $L:y=3$ and passing through the point $P\left(-1,\frac{1}{2}\right)$ is $x=-1$ .