Pablo is researching different types of exercise to see which gives the best calorie burn. He found that an indoor cycling class can burn up to 150 calories every 10 minutes. The elliptical machine burns 270 calories every 30 minutes. Do these two forms of exercise burn the same amount of calories in an hour?

### Equation of Parallel Lines

When two lines are parallel, they have the same slope and never intersect. So, if a given line has a slope of -2, then any line that is parallel to that line will also have a slope of -2, but it will have a different \begin{align*}y-\end{align*}intercept.

Let's find the equation of the following parallel lines.

- Find the equation of the line that is parallel to \begin{align*}y = \frac{2}{3}x -5\end{align*} and passes through (-12, 1).

We know that the slopes will be the same; however we need to find the \begin{align*}y-\end{align*}intercept for this new line. Use the point you were given, (-12, 1) and plug it in for \begin{align*}x\end{align*} and \begin{align*}y\end{align*} to solve for \begin{align*}b\end{align*}.

\begin{align*}y &= \frac{2}{3}x + b\\ 1 &= \frac{2}{3}(-12) + b \\ 1 &= -8 + b\\ 9 &= b\end{align*}

The equation of the parallel line is \begin{align*}y = \frac{2}{3}x + 9\end{align*}.

- Find the equation of the line that passes through (4, -7) and is parallel to \begin{align*}y = -2\end{align*}.

The line \begin{align*}y = -2\end{align*} does not have a \begin{align*}x-\end{align*}term, meaning it has no slope. This is a horizontal line. Therefore, to find the horizontal line that passes through (4, -7), we only need the \begin{align*}y-\end{align*}coordinate. The line would be \begin{align*}y = -7\end{align*}.

The same would be true for vertical lines, but all vertical line equations are in the form \begin{align*}x = a\end{align*}. The \begin{align*}x-\end{align*}coordinate of a given point would be what is needed to determine the equation of the parallel vertical line.

- Find the equation of the line that passes through (6, -10) and is parallel to the line that passes through (4, -6) and (3, -4).

First, we need to find the slope of the line that our line will be parallel to. Use the points (4, -6) and (3, -4) to find the slope.

\begin{align*}m = \frac{-4-(-6)}{3-4} = \frac{2}{-1} = -2\end{align*}

This is the slope of our given line as well as the parallel line. Use the point (6, -10) to find the \begin{align*}y-\end{align*}intercept of the line that we are trying to find the equation for.

\begin{align*}-10 &= -2(6) + b\\ -10 &= -12 + b \\ 2 &= b\end{align*}

The equation of the line is \begin{align*}y = -2x + 2\end{align*}.

### Examples

#### Example 1

Earlier, you were asked if the two forms of exercise burn the same amount of calories in an hour.

The equation for indoor cycling is \begin{align*}y=\frac{150}{10}x\end{align*} or \begin{align*}y=15x\end{align*}. For the elliptical, the equation would be \begin{align*}y=\frac{270}{30}x\end{align*} or \begin{align*}y=9x\end{align*}. The two lines are not parallel, so the two exercises do not burn an equal number of calories. After 60 minutes, indoor cycling will burn 900 calories and the elliptical will burn 540 calories.

#### Example 2

Find the equation of the line that is parallel to \begin{align*}x - 2y = 8\end{align*} and passes through (4, -3).

First, we need to change this line from standard form to slope-intercept form.

\begin{align*}x - 2y &= 8\\ -2y &= -x + 8 \quad \text{Now, we know the slope is} \ \frac{1}{2}. \ \text{Let's find the new} \ y- \ \text{intercept.}\\ y &= \frac{1}{2}x - 4\end{align*}

\begin{align*}-3 &= \frac{1}{2}(4) + b\\ -3 &= 2 + b\\ -5 &= b\end{align*}

The equation of the parallel line is \begin{align*}y = \frac{1}{2}x - 5\end{align*} or \begin{align*}x - 2y = 10\end{align*}.

#### Example 3

Find the equation of the line that is parallel to \begin{align*}x = 9\end{align*} and passes through (-1, 3).

\begin{align*}x = 9\end{align*} is a vertical line that passes through the \begin{align*}x-\end{align*}axis at 9. Therefore, we only need to \begin{align*}x-\end{align*}coordinate of the point to determine the equation of the parallel vertical line. The parallel line through (-1, 3) would be \begin{align*}x = -1\end{align*}.

#### Example 4

Find the equation of the line that passes through (-5, 2) and is parallel to the line that passes through (6, -1) and (1, 3).

First, find the slope between (6, -1) and (1, 3).

\begin{align*}m = \frac{-1-3}{6-1} = \frac{-4}{5} = - \frac{4}{5}\end{align*}

This will also be the slope of the parallel line. Use this slope with the given point, (-5, 2).

\begin{align*}2 &= - \frac{4}{5}(-5) + b\\ 2 &= 1 + b \\ 1 &= b\end{align*}

The equation of the parallel line is \begin{align*}y = - \frac{4}{5}x + 1\end{align*}.

### Review

Find the equation of the line, given the following information. You may leave your answer in slope-intercept form.

- Passes through (4, 7) and is parallel to \begin{align*}x - y = -5\end{align*}.
- Passes through (-6, -2) and is parallel to \begin{align*}y = 4\end{align*}.
- Passes through (-3, 5) and is parallel to \begin{align*}y = - \frac{1}{3}x - 1\end{align*}.
- Passes through (1, -9) and is parallel to \begin{align*}x = 8\end{align*}.
- Passes through the \begin{align*}y-\end{align*}intercept of \begin{align*}2x - 3y = 6\end{align*} and parallel to \begin{align*}x - 4y = 10\end{align*}.
- Passes through (-12, 4) and is parallel to \begin{align*}y = -3x + 5\end{align*}.
- Passes through the \begin{align*}x-\end{align*}intercept of \begin{align*}2x - 3y = 6\end{align*} and parallel to \begin{align*}x + 4y = -3\end{align*}.
- Passes through (7, -8) and is parallel to \begin{align*}2x + 5y = 14\end{align*}.
- Passes through (1, 3) and is parallel to the line that passes through (-6, 2) and (-4, 6).
- Passes through (-18, -10) and is parallel to the line that passes through (-2, 2) and (-8, 1).
- Passes through (-4, -1) and is parallel to the line that passes through (15, 7) and (-1, -1).

Are the pairs of lines parallel? Briefly explain how you know.

- \begin{align*}x - 2y = 4\end{align*} and \begin{align*}-5x + 10y = 16\end{align*}
- \begin{align*}3x + 4y = -8\end{align*} and \begin{align*}6x + 12y = -1\end{align*}
- \begin{align*}5x - 5y = 20\end{align*} and \begin{align*}x + y = 7\end{align*}
- \begin{align*}8x - 12y = 36\end{align*} and \begin{align*}10x - 15y = -15\end{align*}

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 2.4.