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Algebra 1 : Functions and Lines

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Example Questions

Example Question #2 : Perpendicular Lines

Which of the following lines could be perpendicular to the following:

$y=\frac{3}{4}x-2$ $\displaystyle y=\frac{3}{4}x-2$

Possible Answers:

None of the available answers

$y=\frac{3}{4}x+2$ $\displaystyle y=\frac{3}{4}x+2$

$y=\frac{4}{6}x-17$ $\displaystyle y=\frac{4}{6}x-17$

y=0 $\displaystyle y=0$

$y=-1\frac{1}{3}x-128$ $\displaystyle y=-1\frac{1}{3}x-128$

Correct answer:

$y=-1\frac{1}{3}x-128$ $\displaystyle y=-1\frac{1}{3}x-128$

Explanation:

The only marker for whether lines are perpendicular is whether their slopes are the opposite-reciprocal for the other line's slope. The $\displaystyle y$-intercept is not important. Therefore, the line perpendicular to $y=\frac{3}{4}x-2$ $\displaystyle y=\frac{3}{4}x-2$ will have a slope of $m=-\frac{4}{3}$ $\displaystyle m=-\frac{4}{3}$ or $m=-1\frac{1}{3}$ $\displaystyle m=-1\frac{1}{3}$

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Example Question #2 : Equations Of Lines

Find a line perpendicular to the line with the equation:

y=-8x+1 $\displaystyle y=-8x+1$

Possible Answers:

$y=-\frac{1}{8}x-1$ $\displaystyle y=-\frac{1}{8}x-1$

x=8x+2 $\displaystyle x=8x+2$

$y=\frac{1}{8}x-19$ $\displaystyle y=\frac{1}{8}x-19$

y=-8x-1 $\displaystyle y=-8x-1$

Correct answer:

$y=\frac{1}{8}x-19$ $\displaystyle y=\frac{1}{8}x-19$

Explanation:

Lines can be written in the slope-intercept format:

y=mx+b $\displaystyle y=mx+b$

In this format, $\displaystyle m$ equals the line's slope and $\displaystyle b$ represents where the line intercepts the y-axis.

In the given equation:

m=-8 $\displaystyle m=-8$

Perpendicular lines have slopes that are negative reciprocals of each other.

First, we need to find its reciprocal.

$-8\rightarrow-\frac{1}{8}$ $\displaystyle -8\rightarrow-\frac{1}{8}$

Second, we need to rewrite it with the opposite sign.

$-\frac{1}{8}\rightarrow\frac{1}{8}$ $\displaystyle -\frac{1}{8}\rightarrow\frac{1}{8}$

Only one of the choices has a slope of $\frac{1}{8}$ $\displaystyle \frac{1}{8}$.

$y=\frac{1}{8}x-19$ $\displaystyle y=\frac{1}{8}x-19$

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Example Question #2 : Equations Of Lines

Find a line perpendicular to the line with the equation:

$y=\frac{5}{6}x-1$ $\displaystyle y=\frac{5}{6}x-1$

Possible Answers:

$y=\frac{6}{5}x-78$ $\displaystyle y=\frac{6}{5}x-78$

$y=-\frac{5}{6}x$ $\displaystyle y=-\frac{5}{6}x$

$y=\frac{5}{6}x-8$ $\displaystyle y=\frac{5}{6}x-8$

$y=-\frac{6}{5}x+9$ $\displaystyle y=-\frac{6}{5}x+9$

Correct answer:

$y=-\frac{6}{5}x+9$ $\displaystyle y=-\frac{6}{5}x+9$

Explanation:

Lines can be written in the slope-intercept format:

y=mx+b $\displaystyle y=mx+b$

In this format, $\displaystyle m$ equals the line's slope and $\displaystyle b$ represents where the line intercepts the y-axis.

In the given equation:

$m=\frac{5}{6}$ $\displaystyle m=\frac{5}{6}$

Perpendicular lines have slopes that are negative reciprocals of each other.

First, we need to find its reciprocal.

$\frac{5}{6}\rightarrow\frac{6}{5}$ $\displaystyle \frac{5}{6}\rightarrow\frac{6}{5}$

Second, we need to rewrite it with the opposite sign.

$\frac{6}{5}\rightarrow-\frac{6}{5}$ $\displaystyle \frac{6}{5}\rightarrow-\frac{6}{5}$

Only one of the choices has a slope of $-\frac{6}{5}$ $\displaystyle -\frac{6}{5}$.

$y=-\frac{6}{5}x+9$ $\displaystyle y=-\frac{6}{5}x+9$

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Example Question #1 : Perpendicular Lines

Find a line perpendicular to the line with the equation:

$y=-\frac{1}{15}x-12$ $\displaystyle y=-\frac{1}{15}x-12$

Possible Answers:

$y=-\frac{1}{15}x+6$ $\displaystyle y=-\frac{1}{15}x+6$

y=15x+6 $\displaystyle y=15x+6$

$\displaystyle y=-15x+6$

$y=\frac{1}{15}x+6$ $\displaystyle y=\frac{1}{15}x+6$

Correct answer:

y=15x+6 $\displaystyle y=15x+6$

Explanation:

Lines can be written in the slope-intercept format:

y=mx+b $\displaystyle y=mx+b$

In this format, $\displaystyle m$ equals the line's slope and $\displaystyle b$ represents where the line intercepts the y-axis.

In the given equation:

$m=-\frac{1}{15}$ $\displaystyle m=-\frac{1}{15}$

Perpendicular lines have slopes that are negative reciprocals of each other.

First, we need to find its reciprocal.

$-\frac{1}{15}\rightarrow-\frac{15}{1}$ $\displaystyle -\frac{1}{15}\rightarrow-\frac{15}{1}$

Rewrite.

$-\frac{15}{1}=-15$ $\displaystyle -\frac{15}{1}=-15$

Second, we need to rewrite it with the opposite sign.

$-15\rightarrow15$ $\displaystyle -15\rightarrow15$

Only one of the choices has a slope of $\displaystyle 15$.

y=15x+6 $\displaystyle y=15x+6$

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Example Question #2 : Perpendicular Lines

Find a line perpendicular to the line with the equation:

$y=\frac{1}{2}x+79$ $\displaystyle y=\frac{1}{2}x+79$

Possible Answers:

y=2x-10 $\displaystyle y=2x-10$

$y=-\frac{1}{2}x-10$ $\displaystyle y=-\frac{1}{2}x-10$

$\displaystyle y=-2x-10$

$y=\frac{1}{2}x-10$ $\displaystyle y=\frac{1}{2}x-10$

Correct answer:

$\displaystyle y=-2x-10$

Explanation:

Lines can be written in the slope-intercept format:

y=mx+b $\displaystyle y=mx+b$

In this format, $\displaystyle m$ equals the line's slope and $\displaystyle b$ represents where the line intercepts the y-axis.

In the given equation:

$m=\frac{1}{2}$ $\displaystyle m=\frac{1}{2}$

Perpendicular lines have slopes that are negative reciprocals of each other.

First, we need to find its reciprocal.

$\frac{1}{2}\rightarrow\frac{2}{1}$ $\displaystyle \frac{1}{2}\rightarrow\frac{2}{1}$

Rewrite.

$\frac{2}{1}=2$ $\displaystyle \frac{2}{1}=2$

Second, we need to rewrite it with the opposite sign.

$2\rightarrow-2$ $\displaystyle 2\rightarrow-2$

Only one of the choices has a slope of $\displaystyle -2$.

$\displaystyle y=-2x-10$

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Example Question #11 : How To Find Out If Lines Are Perpendicular

Find a line perpendicular to the line with the equation:

$y=\frac{7}{9}x-1$ $\displaystyle y=\frac{7}{9}x-1$

Possible Answers:

$y=-\frac{7}{9}x-8$ $\displaystyle y=-\frac{7}{9}x-8$

$y=\frac{9}{7}x+15$ $\displaystyle y=\frac{9}{7}x+15$

$y=\frac{7}{9}x+6$ $\displaystyle y=\frac{7}{9}x+6$

$y=-\frac{9}{7}x+9$ $\displaystyle y=-\frac{9}{7}x+9$

Correct answer:

$y=-\frac{9}{7}x+9$ $\displaystyle y=-\frac{9}{7}x+9$

Explanation:

Lines can be written in the slope-intercept format:

y=mx+b $\displaystyle y=mx+b$

In this format, $\displaystyle m$ equals the line's slope and $\displaystyle b$ represents where the line intercepts the y-axis.

In the given equation:

$m=\frac{7}{9}$ $\displaystyle m=\frac{7}{9}$

Perpendicular lines have slopes that are negative reciprocals of each other.

First, we need to find its reciprocal.

$\frac{7}{9}\rightarrow\frac{9}{7}$ $\displaystyle \frac{7}{9}\rightarrow\frac{9}{7}$

Second, we need to rewrite it with the opposite sign.

$\frac{9}{7}\rightarrow-\frac{9}{7}$ $\displaystyle \frac{9}{7}\rightarrow-\frac{9}{7}$

Only one of the choices has a slope of $-\frac{9}{7}$ $\displaystyle -\frac{9}{7}$.

$y=-\frac{9}{7}x+9$ $\displaystyle y=-\frac{9}{7}x+9$

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Example Question #12 : Perpendicular Lines

Find a line perpendicular to the line with the equation:

$y=\frac{5}{2}x-2$ $\displaystyle y=\frac{5}{2}x-2$

Possible Answers:

y=-5x-7 $\displaystyle y=-5x-7$

$y=-\frac{2}{5}x+9$ $\displaystyle y=-\frac{2}{5}x+9$

$y=\frac{2}{5}x+1$ $\displaystyle y=\frac{2}{5}x+1$

$y=\frac{5}{2}x-8$ $\displaystyle y=\frac{5}{2}x-8$

Correct answer:

$y=-\frac{2}{5}x+9$ $\displaystyle y=-\frac{2}{5}x+9$

Explanation:

Lines can be written in the slope-intercept format:

y=mx+b $\displaystyle y=mx+b$

In this format, $\displaystyle m$ equals the line's slope and $\displaystyle b$ represents where the line intercepts the y-axis.

In the given equation:

$m=\frac{5}{2}$ $\displaystyle m=\frac{5}{2}$

Perpendicular lines have slopes that are negative reciprocals of each other.

First, we need to find its reciprocal.

$\frac{5}{2}\rightarrow\frac{2}{5}$ $\displaystyle \frac{5}{2}\rightarrow\frac{2}{5}$

Second, we need to rewrite it with the opposite sign.

$\frac{2}{5}\rightarrow-\frac{2}{5}$ $\displaystyle \frac{2}{5}\rightarrow-\frac{2}{5}$

Only one of the choices has a slope of $-\frac{2}{5}$ $\displaystyle -\frac{2}{5}$.

$y=-\frac{2}{5}x+9$ $\displaystyle y=-\frac{2}{5}x+9$

Report an Error

Example Question #13 : Perpendicular Lines

Find a line perpendicular to the line with the equation:

$y=-\frac{5}{7}x-61$ $\displaystyle y=-\frac{5}{7}x-61$

Possible Answers:

$y=\frac{7}{5}x+159$ $\displaystyle y=\frac{7}{5}x+159$

$y=-\frac{7}{5}x-78$ $\displaystyle y=-\frac{7}{5}x-78$

$y=\frac{7}{6}x+78$ $\displaystyle y=\frac{7}{6}x+78$

$y=\frac{5}{7}x+97$ $\displaystyle y=\frac{5}{7}x+97$

Correct answer:

$y=\frac{7}{5}x+159$ $\displaystyle y=\frac{7}{5}x+159$

Explanation:

Lines can be written in the slope-intercept format:

y=mx+b $\displaystyle y=mx+b$

In this format, $\displaystyle m$ equals the line's slope and $\displaystyle b$ represents where the line intercepts the y-axis.

In the given equation:

$m=-\frac{5}{7}$ $\displaystyle m=-\frac{5}{7}$

Perpendicular lines have slopes that are negative reciprocals of each other.

First, we need to find its reciprocal.

$-\frac{5}{7}\rightarrow-\frac{7}{5}$ $\displaystyle -\frac{5}{7}\rightarrow-\frac{7}{5}$

Second, we need to rewrite it with the opposite sign.

$-\frac{7}{5}\rightarrow\frac{7}{5}$ $\displaystyle -\frac{7}{5}\rightarrow\frac{7}{5}$

Only one of the choices has a slope of $\frac{7}{5}$ $\displaystyle \frac{7}{5}$.

$y=\frac{7}{5}x+159$ $\displaystyle y=\frac{7}{5}x+159$

Report an Error

Example Question #14 : Perpendicular Lines

Find a line perpendicular to the line with the equation:

$y=\frac{3}{4}x-23$ $\displaystyle y=\frac{3}{4}x-23$

Possible Answers:

$y=-\frac{3}{4}x+84$ $\displaystyle y=-\frac{3}{4}x+84$

$y=\frac{4}{3}x-88$ $\displaystyle y=\frac{4}{3}x-88$

$y=\frac{3}{4}x+56$ $\displaystyle y=\frac{3}{4}x+56$

$y=-\frac{4}{3}x+87$ $\displaystyle y=-\frac{4}{3}x+87$

Correct answer:

$y=-\frac{4}{3}x+87$ $\displaystyle y=-\frac{4}{3}x+87$

Explanation:

Lines can be written in the slope-intercept format:

y=mx+b $\displaystyle y=mx+b$

In this format, $\displaystyle m$ equals the line's slope and $\displaystyle b$ represents where the line intercepts the y-axis.

In the given equation:

$m=\frac{3}{4}$ $\displaystyle m=\frac{3}{4}$

Perpendicular lines have slopes that are negative reciprocals of each other.

First, we need to find its reciprocal.

$\frac{3}{4}\rightarrow\frac{4}{3}$ $\displaystyle \frac{3}{4}\rightarrow\frac{4}{3}$

Second, we need to rewrite it with the opposite sign.

$\frac{4}{3}\rightarrow-\frac{4}{3}$ $\displaystyle \frac{4}{3}\rightarrow-\frac{4}{3}$

Only one of the choices has a slope of $-\frac{4}{3}$ $\displaystyle -\frac{4}{3}$.

$y=-\frac{4}{3}x+87$ $\displaystyle y=-\frac{4}{3}x+87$

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Example Question #15 : Perpendicular Lines

Find a line perpendicular to the line with the equation:

y=2x-10 $\displaystyle y=2x-10$

Possible Answers:

y=-2x+2 $\displaystyle y=-2x+2$

y=2x+2 $\displaystyle y=2x+2$

$y=-\frac{1}{2}$ $\displaystyle y=-\frac{1}{2}$

$y=-\frac{1}{2}x+2$ $\displaystyle y=-\frac{1}{2}x+2$

Correct answer:

$y=-\frac{1}{2}x+2$ $\displaystyle y=-\frac{1}{2}x+2$

Explanation:

Lines can be written in the slope-intercept format:

y=mx+b $\displaystyle y=mx+b$

In this format, $\displaystyle m$ equals the line's slope and $\displaystyle b$ represents where the line intercepts the y-axis.

In the given equation:

m=2 $\displaystyle m=2$

Perpendicular lines have slopes that are negative reciprocals of each other.

First, we need to find its reciprocal.

$2\rightarrow\frac{1}{2}$ $\displaystyle 2\rightarrow\frac{1}{2}$

Second, we need to rewrite it with the opposite sign.

$\frac{1}{2}\rightarrow-\frac{1}{2}$ $\displaystyle \frac{1}{2}\rightarrow-\frac{1}{2}$

Only one of the choices has a slope of $-\frac{1}{2}$ $\displaystyle -\frac{1}{2}$.

$y=-\frac{1}{2}x+2$ $\displaystyle y=-\frac{1}{2}x+2$

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Algebra 1 : Functions and Lines

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #2 : Perpendicular Lines

Example Question #2 : Equations Of Lines

Example Question #2 : Equations Of Lines

Example Question #1 : Perpendicular Lines

Example Question #2 : Perpendicular Lines

Example Question #11 : How To Find Out If Lines Are Perpendicular

Example Question #12 : Perpendicular Lines

Example Question #13 : Perpendicular Lines

Example Question #14 : Perpendicular Lines

Example Question #15 : Perpendicular Lines

All Algebra 1 Resources

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