# 4.5 Use the slope–intercept form of an equation of a line  (Page 7/12)

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Use slopes and y -intercepts to determine if the lines $x=1$ and $x=-5$ are parallel.

parallel

Use slopes and y -intercepts to determine if the lines $x=8$ and $x=-6$ are parallel.

parallel

Use slopes and y -intercepts to determine if the lines $y=2x-3$ and $-6x+3y=-9$ are parallel. You may want to graph these lines, too, to see what they look like.

## Solution

$\begin{array}{cccccccc}\begin{array}{}\\ \text{The first equation is already in slope–intercept form.}\hfill \\ \text{Solve the second equation for}\phantom{\rule{0.2em}{0ex}}y.\hfill \\ \\ \\ \\ \end{array}\hfill & & & \phantom{\rule{1.5em}{0ex}}\begin{array}{ccc}\hfill y& =\hfill & 2x-3\hfill \\ \hfill y& =\hfill & 2x-3\hfill \\ \hfill -6x+3y& =\hfill & -9\hfill \\ \hfill 3y& =\hfill & 6x-9\hfill \\ \hfill \frac{3y}{3}& =\hfill & \frac{6x-9}{3}\hfill \end{array}\hfill & \hfill \begin{array}{c}\hfill \text{and}\hfill \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{array}\hfill & & & \begin{array}{c}-6x+3y=-9\hfill \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{array}\hfill \\ \begin{array}{}\\ \\ \\ \\ \text{The second equation is now in}\hfill \\ \text{slope–intercept form.}\hfill \\ \text{Identify the slope and}\phantom{\rule{0.2em}{0ex}}y\text{-intercept of both lines.}\hfill \\ \\ \\ \end{array}\hfill & & & \phantom{\rule{4.5em}{0ex}}\begin{array}{ccc}\hfill y& =\hfill & 2x-3\hfill \\ \hfill y& =\hfill & 2x-3\hfill \\ \hfill y& =\hfill & mx+b\hfill \\ \hfill m& =\hfill & 2\hfill \end{array}\hfill & & & & \begin{array}{}\\ \\ \\ \\ \\ \\ \hfill y& =\hfill & 2x-3\hfill \\ \hfill y& =\hfill & mx+b\hfill \\ \hfill m& =\hfill & 2\hfill \end{array}\hfill \\ & & & \phantom{\rule{3.5em}{0ex}}y\text{-intercept is (0 ,−3)}\hfill & & & & y\text{-intercept is (0, −3)}\hfill \end{array}$

The lines have the same slope, but they also have the same y -intercepts. Their equations represent the same line. They are not parallel; they are the same line.

Use slopes and y -intercepts to determine if the lines $y=-\frac{1}{2}x-1$ and $x+2y=2$ are parallel.

not parallel; same line

Use slopes and y -intercepts to determine if the lines $y=\frac{3}{4}x-3$ and $3x-4y=12$ are parallel.

not parallel; same line

## Use slopes to identify perpendicular lines

Let’s look at the lines whose equations are $y=\frac{1}{4}x-1$ and $y=-4x+2$ , shown in [link] .

These lines lie in the same plane and intersect in right angles. We call these lines perpendicular .

What do you notice about the slopes of these two lines? As we read from left to right, the line $y=\frac{1}{4}x-1$ rises, so its slope is positive. The line $y=-4x+2$ drops from left to right, so it has a negative slope. Does it make sense to you that the slopes of two perpendicular lines will have opposite signs?

If we look at the slope of the first line, ${m}_{1}=\frac{1}{4}$ , and the slope of the second line, ${m}_{2}=-4$ , we can see that they are negative reciprocals of each other. If we multiply them, their product is $-1.$

$\begin{array}{c}{m}_{1}·{m}_{2}\hfill \\ \frac{1}{4}\left(-4\right)\hfill \\ -1\hfill \end{array}$

This is always true for perpendicular lines    and leads us to this definition.

## Perpendicular lines

Perpendicular lines are lines in the same plane that form a right angle.

If ${m}_{1}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{m}_{2}$ are the slopes of two perpendicular lines, then:

${m}_{1}·{m}_{2}=-1\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}{m}_{1}=\frac{-1}{{m}_{2}}$

Vertical lines and horizontal lines are always perpendicular to each other.

We were able to look at the slope–intercept form of linear equations and determine whether or not the lines were parallel. We can do the same thing for perpendicular lines.

We find the slope–intercept form of the equation, and then see if the slopes are negative reciprocals. If the product of the slopes is $-1$ , the lines are perpendicular. Perpendicular lines may have the same y -intercepts.

Use slopes to determine if the lines, $y=-5x-4$ and $x-5y=5$ are perpendicular.

## Solution

$\begin{array}{ccccccccccc}\text{The first equation is in slope–intercept form.}\hfill & & & \hfill \phantom{\rule{2em}{0ex}}y& =\hfill & -5x-4\hfill & & & & & \\ \text{Solve the second equation for}\phantom{\rule{0.2em}{0ex}}y.\hfill & & & \hfill \phantom{\rule{2em}{0ex}}x-5y& =\hfill & 5\hfill & & & & & \\ & & & \hfill \phantom{\rule{2em}{0ex}}-5y& =\hfill & \text{−}x+5\hfill & & & & & \\ & & & \hfill \phantom{\rule{2em}{0ex}}\frac{-5y}{-5}& =\hfill & \frac{\text{−}x+5}{-5}\hfill & & & & & \\ & & & \hfill \phantom{\rule{2em}{0ex}}y& =\hfill & \frac{1}{5}x-1\hfill & & & & & \\ \text{Identify the slope of each line.}\hfill & & & \hfill \phantom{\rule{2em}{0ex}}y& =\hfill & -5x-4\hfill & & & & & \hfill y& =\hfill & \frac{1}{5}x-1\hfill \\ & & & \hfill \phantom{\rule{2em}{0ex}}y& =\hfill & mx+b\hfill & & & & & \hfill y& =\hfill & mx+b\hfill \\ & & & \hfill \phantom{\rule{2em}{0ex}}{m}_{1}& =\hfill & -5\hfill & & & & & \hfill {m}_{2}& =\hfill & \frac{1}{5}\hfill \end{array}$

The slopes are negative reciprocals of each other, so the lines are perpendicular. We check by multiplying the slopes,

$\begin{array}{c}{m}_{1}·{m}_{2}\hfill \\ \\ -5\left(\frac{1}{5}\right)\hfill \\ -1✓\hfill \end{array}$

Use slopes to determine if the lines $y=-3x+2$ and $x-3y=4$ are perpendicular.

perpendicular

Use slopes to determine if the lines $y=2x-5$ and $x+2y=-6$ are perpendicular.

perpendicular

Use slopes to determine if the lines, $7x+2y=3$ and $2x+7y=5$ are perpendicular.

## Solution

$\begin{array}{ccccccccccccccc}\text{Solve the equations for}\phantom{\rule{0.2em}{0ex}}y.\hfill & & & & & \hfill \phantom{\rule{2em}{0ex}}7x+2y& =\hfill & 3\hfill & & & & & \hfill 2x+7y& =\hfill & 5\hfill \\ & & & & & \hfill \phantom{\rule{2em}{0ex}}2y& =\hfill & -7x+3\hfill & & & & & \hfill 7y& =\hfill & -2x+5\hfill \\ & & & & & \hfill \phantom{\rule{2em}{0ex}}\frac{2y}{2}& =\hfill & \frac{-7x+3}{2}\hfill & & & & & \hfill \frac{7y}{7}& =\hfill & \frac{-2x+5}{7}\hfill \\ & & & & & \hfill \phantom{\rule{2em}{0ex}}y& =\hfill & -\frac{7}{2}x+\frac{3}{2}\hfill & & & & & \hfill y& =\hfill & -\frac{2}{7}x+\frac{5}{7}\hfill \\ \text{Identify the slope of each line.}\hfill & & & & & \hfill \phantom{\rule{2em}{0ex}}y& =\hfill & mx+b\hfill & & & & & \hfill y& =\hfill & mx+b\hfill \\ & & & & & \hfill \phantom{\rule{2em}{0ex}}{m}_{1}& =\hfill & -\frac{7}{2}\hfill & & & & & \hfill {m}_{2}& =\hfill & -\frac{2}{7}\hfill \end{array}$

The slopes are reciprocals of each other, but they have the same sign. Since they are not negative reciprocals, the lines are not perpendicular.

4x+7y=29,x+3y=11 substitute method of linear equation
substitute method of linear equation
Srinu
Solve one equation for one variable. Using the 2nd equation, x=11-3y. Substitute that for x in first equation. this will find y. then use the value for y to find the value for x.
bruce
I want to learn
Elizebeth
help
Elizebeth
I want to learn. Please teach me?
Wayne
1) Use any equation, and solve for any of the variables. Since the coefficient of x (the number in front of the x) in the second equation is 1 (it actually isn't shown, but 1 * x = x), use that equation. Subtract 3y from both sides (this isolates the x on the left side of the equal sign).
bruce
2) This results in x=11-3y. x is note in terms of y. Use that as the value of x and substitute for all x in the first equation. The first equation becomes 4(11-3y)+7y =29. Note that the only variable left in the first equation is the y. If you have multiple variable, then something is wrong.
bruce
3) Distribute (multiply) the 4 across 11-3y to get 44-12y. Add this to the 7y. So, the equation is now 44-5y=29.
bruce
4) Solve 44-5y=29 for y. Isolate the y by subtracting 44 from birth sides, resulting in -5y=-15. Now, divide birth sides by -5 (since you have -5y). This results in y=3. You now have the value of one variable.
bruce
5) The last step is to take the value of y from Step 4) and substitute into the 2nd equation. Therefore: x+3y=11 becomes x+3(3)=11. Then multiplying, x+9=11. Finally, solve for x by subtracting 9 from both sides. Therefore, x=2.
bruce
6) The ordered pair of (2, 3) is the proposed solution. To check, substitute those values into either equation. If the result is true, then the solution is correct. 4(2)+7(3)=8+21=29. TRUE! Finished.
bruce
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Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike?
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Susanna if the first cooler holds five times the gallons then the other cooler. The big cooler holda 40 gallons and the 2nd will hold 8 gallons is that correct?
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@Susanna that person is correct if you divide 40 by 8 you can see it's 5 it's simple
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