Call Now to Set Up Tutoring:

(888) 888-0446

Algebra 1 : Functions and Lines

Study concepts, example questions & explanations for Algebra 1

Create An Account

All Algebra 1 Resources

10 Diagnostic Tests 557 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #16 : Perpendicular Lines

Find a line perpendicular to the line with the equation:

$y=-5x+\frac{1}{5}$

Possible Answers:

$y=-\frac{1}{5}x-8$

y=5x+2

y=-5x+5

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

Correct answer:

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

Explanation:

Lines can be written in the slope-intercept format:

y=mx+b

In this format, equals the line's slope and represents where the line intercepts the y-axis.

In the given equation:

m=-5

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

First, we need to find its reciprocal.

$-5\rightarrow-\frac{1}{5}$

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

$-\frac{1}{5}\rightarrow\frac{1}{5}$

Only one of the choices has a slope of $\frac{1}{5}$ .

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

Report an Error

Example Question #17 : Perpendicular Lines

Find a line perpendicular to the line with the equation:

$y=\frac{3}{10}x-9$

Possible Answers:

$y=\frac{3}{10}x+100$

$y=-\frac{10}{3}x+98$

$y=\frac{10}{3}x+100$

$y=-\frac{3}{10}x-78$

Correct answer:

$y=-\frac{10}{3}x+98$

Explanation:

Lines can be written in the slope-intercept format:

y=mx+b

In this format, equals the line's slope and represents where the line intercepts the y-axis.

In the given equation:

$m=\frac{3}{10}$

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

First, we need to find its reciprocal.

$\frac{3}{10}\rightarrow\frac{10}{3}$

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

$\frac{10}{3}\rightarrow-\frac{10}{3}$

Only one of the choices has a slope of $-\frac{10}{3}$ .

$y=-\frac{10}{3}x+98$

Report an Error

Example Question #18 : Perpendicular Lines

Find a line perpendicular to the line with the equation:

y=100x-2

Possible Answers:

y=-102x-2

$y=-\frac{1}{100}x+85$

$y=\frac{1}{100}x+55$

y=100x+102

Correct answer:

$y=-\frac{1}{100}x+85$

Explanation:

Lines can be written in the slope-intercept format:

y=mx+b

In this format, equals the line's slope and represents where the line intercepts the y-axis.

In the given equation:

m=100

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

First, we need to find its reciprocal.

$100\rightarrow\frac{1}{100}$

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

$\frac{1}{100}\rightarrow-\frac{1}{100}$

Only one of the choices has a slope of $-\frac{1}{100}$ .

$y=-\frac{1}{100}x+85$

Report an Error

Example Question #19 : Perpendicular Lines

Given the two lines: y=2x and $y= -\frac{1}{2}x$ , are the lines perpendicular to each other?

Possible Answers:

$\textup{Yes, because both lines are in slope-intercept form.}$

$\textup{No, the slopes must be the same for both lines.}$

$\textup{Yes, the slopes are negative reciprocals of each other.}$

$\textup{No, because the slopes have no y-intercept.}$

$\textup{No, the slopes are not negative reciprocals of each other.}$

Correct answer:

$\textup{Yes, the slopes are negative reciprocals of each other.}$

Explanation:

Write the perpendicular line slope formula. The perpendicular slope is the negative reciprocal of the original slope.

$m_{perpendicular} = -\frac{1}{m_{original}}$

Let y=2x be the original equation. The slope is . Substitute this into the equation to find the slope of any perpendicular line.

$m_{perpendicular} = -\frac{1}{2}$

The slope of a perpendicular line must have a slope of $-\frac{1}{2}$ , which is also the slope for $y= -\frac{1}{2}x$ .

The answer is:

$\textup{Yes, the slopes are negative reciprocal to each other.}$

Report an Error

Example Question #3541 : Algebra 1

Select the equation of the line that is perpendicular to $y=\frac{1}{2}x-4$ .

Possible Answers:

None of the other answers.

y=-2x+3

$y=-\frac{1}{2}x+3$

y=2x+3

y=2x-4

Correct answer:

y=-2x+3

Explanation:

Lines are perpendicular if their slopes are negative reciprocals of one another. For example, the negative reciprocal of $3=-\frac{1}{3}$ . So y=-2x+3 is perpendicular to $y=\frac{1}{2}x-4$ because their slopes are the negative reciprocals of each other. $\frac{1}{2}=-2\hspace{1mm}and\hspace{1mm} -2=-\frac{1}{-2}=\frac{1}{2}$ . A positive slope can still be the negative reciprocal as you can see.

Report an Error

Example Question #261 : Functions And Lines

Are the lines y=-3x-3 and $y=-\frac{1}{3}x+\frac{1}{3}$ perpendicular to each other? If so, why?

Possible Answers:

No, because the slopes are not the negative reciprocal of each other.

Yes, because the slopes are the negative reciprocal of each other.

No, because the functions are not the negative reciprocal to each other.

Yes, because the slopes are the reciprocal of each other.

No, because the slopes are not the reciprocal of each other.

Correct answer:

No, because the slopes are not the negative reciprocal of each other.

Explanation:

The equations are already in slope-intercept form: y=mx+b . In order to determine whether the slopes are perpendicular to each other, write the formula for the slope of the perpendicular line.

$m_{\textup{perpendicular}} = -\frac{1}{m_{\textup{original}}}$

The perpendicular slope is the negative reciprocal of the original slope. Take an original equation, perhaps y=-3x-3 .

The slope of the original equation is . Substitute this in the equation.

$m_{\textup{perpendicular}} = -\frac{1}{-3} = \frac{1}{3}$

The perpendicular line must have a slope of $\frac{1}{3}$ . Since this does not match the slope of $y=-\frac{1}{3}x+\frac{1}{3}$ , the lines are NOT perpendicular to each other.

The answer is:

$\textup{No, because the slopes are not the negative reciprocal of each other.}$

Report an Error

Example Question #22 : How To Find Out If Lines Are Perpendicular

Which of the following equations is perpendicular to the given function:

y=4x+15

Possible Answers:

$y=\frac{-1}{2}x+35$

y=4x-3

$y=\frac{-1}{4}x+365$

$y=\frac{1}{4}x+17$

Correct answer:

$y=\frac{-1}{4}x+365$

Explanation:

Which of the following equations is perpendicular to the given function:

y=4x+15

To find a line perpendicular to a given linear function, simply find the opposit reciprocal of the slope of the given function.

So, we begin with 4, the opposite reciprocal will have the opposite sign and will be the flipped fraction of 4, so it will look like this:

$\frac{-1}{4}$

So we need to choose the answer with the correct slope, choose the only option with slope:

$\frac{-1}{4}$

Report an Error

Example Question #23 : How To Find Out If Lines Are Perpendicular

Which line is perpendicular to the following line:

y = 4x + 7

Possible Answers:

2y = 8x + 9

-2y = 4x - 11

4y = x + 2

-4y = x - 5

y = -4x - 1

Correct answer:

-4y = x - 5

Explanation:

To find a line that is perpendicular to another line, we must look at the slope. Two lines that are perpendicular have slopes that are opposite reciprocals of one another.

Opposite reciprocals mean we change the sign and switch the numerator and denominator. Here is an example:

The reciprocal of is $- \frac{1}{4}$

The reciprocal of $\frac{1}{2}$ is

So, in the given equation of a line,

y = 4x + 7

We see that the slope is . To find a line perpendicular to this one, we need to find a line with a slope of $- \frac{1}{4}$ .

In the following equation,

-4y = x - 5

We must first write it in slope intercept form. To do that, we divide each term by .

$\frac{-4y}{-4} = \frac{x}{-4} - \frac{5}{-4}$

$y = \frac{x}{-4} + \frac{5}{4}$

$y = -\frac{1}{4}x + \frac{5}{4}$

So, the slope of this line is $-\frac{1}{4}$ which makes it perpendicular to the original line.

Report an Error

Example Question #21 : Equations Of Lines

Choose the lines that are perpendicular.

Possible Answers:

y=3x+5

$y=\frac{1}{3}x+3$

None of these

y=7x+4

y=4x+7

y=-2x+4

y=-2x-5

y=5x+2

$y=-\frac{1}{5}x+4$

Correct answer:

y=5x+2

$y=-\frac{1}{5}x+4$

Explanation:

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

For example, the negative reciprocal of 2 is $-\frac{1}{2}$ and the negative reciprocal of $-\frac{1}{5}$ would be 5.

Thus,

y=5x+2

is perpendicular to

$y=-\frac{1}{5}x+4$ .

Report an Error

Example Question #23 : Perpendicular Lines

Find the line that is perpendicular to the following:

-9y = 54x + 27

Possible Answers:

y = 54x - 13

3y = -18x + 12

4y = 24x - 2

-9y = 3x +7

12y = 2x+1

Correct answer:

12y = 2x+1

Explanation:

Two lines are perpendicular if their slopes are opposite reciprocals of each other (opposite: different signs, reciprocal: numerator and denominator are switched).

To find the slopes, we will write the original equation in slope-intercept form

y = mx+b

where m is the slope. Given the original equation

-9y = 54x + 27

we must solve for y. To do that, we will divide each term by -9. We get

$\frac{-9y}{-9} = \frac{54x}{-9} + \frac{27}{-9}$

y = -6x -3

Therefore, the slope of this line is -6. We must find a line that has a slope that is the opposite reciprocal of this line. The opposite reciprocal slope of -6 is $\frac{1}{6}$ .

Let's look at the line

12y = 2x+1

We must write it in slope-intercept form. To do that, we will divide each term by 12. We get

$\frac{12y}{12} = \frac{2x}{12} + \frac{1}{12}$

$y = \frac{2}{12}x + \frac{1}{12}$

We can simplify to

$y = \frac{1}{6}x + \frac{1}{12}$

The slope of this line is $\frac{1}{6}$ . Therefore, it is perpendicular to the original line.

Report an Error

View Algebra Tutors

Grace
Certified Tutor

University of Massachusetts-Dartmouth, Bachelor of Education, Portuguese. University of Massachusetts-Dartmouth, Masters in B...

View Algebra Tutors

Liban
Certified Tutor

Emory University, Bachelor of Science, Mathematics/Economics.

View Algebra Tutors

Joseph
Certified Tutor

Rutgers University-New Brunswick, Bachelor in Arts, Chemistry. Rutgers Graduate School of Biomedical Sciences, Current Grad S...

All Algebra 1 Resources

10 Diagnostic Tests 557 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GRE Tutors in Dallas Fort Worth, Algebra Tutors in San Francisco-Bay Area, LSAT Tutors in Atlanta, GRE Tutors in Denver, Chemistry Tutors in Atlanta, ACT Tutors in Los Angeles, Computer Science Tutors in Dallas Fort Worth, Math Tutors in New York City, SAT Tutors in Chicago, Chemistry Tutors in Phoenix

Popular Courses & Classes

SSAT Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in Philadelphia, LSAT Courses & Classes in Atlanta, SAT Courses & Classes in Phoenix, MCAT Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in Los Angeles, LSAT Courses & Classes in Chicago, MCAT Courses & Classes in Atlanta, ACT Courses & Classes in Chicago, LSAT Courses & Classes in Los Angeles

Popular Test Prep

MCAT Test Prep in Chicago, MCAT Test Prep in Seattle, LSAT Test Prep in New York City, MCAT Test Prep in Atlanta, ACT Test Prep in Chicago, ISEE Test Prep in Denver, SAT Test Prep in Dallas Fort Worth, SAT Test Prep in Philadelphia, LSAT Test Prep in Seattle, LSAT Test Prep in Los Angeles

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Algebra 1 : Functions and Lines

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #16 : Perpendicular Lines

Example Question #17 : Perpendicular Lines

Example Question #18 : Perpendicular Lines

Example Question #19 : Perpendicular Lines

Example Question #3541 : Algebra 1

Example Question #261 : Functions And Lines

Example Question #22 : How To Find Out If Lines Are Perpendicular

Example Question #23 : How To Find Out If Lines Are Perpendicular

Example Question #21 : Equations Of Lines

Example Question #23 : Perpendicular Lines

All Algebra 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: