Call Now to Set Up Tutoring:

(888) 888-0446

Intermediate Geometry : Lines

Study concepts, example questions & explanations for Intermediate Geometry

Create An Account

All Intermediate Geometry Resources

8 Diagnostic Tests 250 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 4 5 6 7 8 9 10 11 … 22 23 Next →

Example Question #6 : How To Find The Length Of A Line With Distance Formula

Find the distance of the line connecting the pair of points

(1, 2) and (4, 6) .

Possible Answers:

Correct answer:

Explanation:

By the distance formula

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2 }$

where (x_1,y_1)=(1,2) and (x_2,y_2)=(4,6)

we have

$\sqrt{(4-1)^2+(6-2)^2 }=\sqrt{(3)^2+(4)^2}= \sqrt{9+16} = \sqrt{25} = 5$

Report an Error

Example Question #7 : How To Find The Length Of A Line With Distance Formula

Find the length of $\small y = \frac{1}{5} x - 2$ for $\small -2 \leq x \leq 5$

Possible Answers:

$\small \sqrt{26.96}$

$\small \small \sqrt{60.56}$

$\small \small \sqrt{50.96}$

$\small \sqrt{36.56}$

$\small \sqrt{9.36}$

Correct answer:

$\small \small \sqrt{50.96}$

Explanation:

To find the distance, first we have to find the specific coordinate pairs that we're finding the distance between. We know the x-values, so to find the y-values we can plug these endpoint x-values into the line:

$\small y = \frac{1}{5}(-2)-2$ first multiply

$\small y = -0.4 - 2$ then subtract

$\small y = -2.4$

$\small y = \frac{1}{5}(5)-2$ first multiply

$\small y = 1 - 2$ then subtract

$\small y = -1$

Now we know that we're finding the distance between the points $\small \small (5, -1)$ and $\small (-2, -2.4)$ . We can just plug these values into the distance formula, using the first pair as $\small (x_{1}, y_{1})$ and the second pair as $\small (x_{2}, y_{2})$ . It would work either way since we are squaring these values, this just makes it easier.

$\small \sqrt{(x_{1}-x_{2})^2 + (y_{1}- y_{2})^2}$

$\small \sqrt{(5--2)^2 + (-1 - -2.4)^2 }$

$\small \sqrt{(5+2)^2 + (-1 + 2.4)^2}$

$\small \sqrt{7^2 + 1.4^2}$

$\small \sqrt{49+1.96} = \sqrt{50.96}$

Report an Error

Example Question #8 : How To Find The Length Of A Line With Distance Formula

Find the length of $\small y = -3x + 4$ for $\small 0 \leq x \leq 5$

Possible Answers:

$5\sqrt{10}$

$4\sqrt{17}$

$\small \small \sqrt{74}$

$\small \sqrt{554}$

Correct answer:

$5\sqrt{10}$

Explanation:

$\small \small y = -3(0)+4$ first multiply

$\small \small y = 0 +4$ then add

$\small \small y = 4$

$\small \small y =-3(5)+4$ first multiply

$\small \small y = -15 +4$ then add

$\small \small y = -11$

Now we know that we're finding the distance between the points $\small (0,4)$ and $\small (5, -11)$ . We can just plug these values into the distance formula, using the first pair as $\small (x_{1}, y_{1})$ and the second pair as $\small (x_{2}, y_{2})$ . Note that it would work either way since we are squaring these values anyway.

$\small \sqrt{(x_{1}-x_{2})^2 + (y_{1}- y_{2})^2}$

$\small \small \sqrt{(0-5)^2 + (4 - -11)^2 }$

$\small \small \sqrt{(-5)^2 + (15)^2}$

$\small \small \sqrt{25 + 225} = \sqrt{250}$

Report an Error

Example Question #9 : How To Find The Length Of A Line With Distance Formula

Find the length of the line $\small y= \frac{1}{2}x - 6$ for $\small -1 \leq y \leq 8$

Possible Answers:

$\small \sqrt{137.25}$

$\small \sqrt{733}$

$9\sqrt{5}$

$\small \sqrt{481}$

$\small \sqrt{373}$

Correct answer:

$9\sqrt{5}$

Explanation:

To find the distance, first we have to find the specific coordinate pairs that we're finding the distance between. We know the y-values, so to find the x-values we can plug these endpoint y-values into the line:

$\small -1 = \frac{1}{2}x - 6$ add 6 to both sides

$\small 5 = \frac{1}{2} x$ multiply by 2

10 = x this endpoint is (10, -1)

$\small 8 = \frac{1}{2} x -6$ add 6 to both sides

$\small 14 = \frac{1}{2}x$ multiply by 2

$\small 28 = x$ this endpoint is (28, 8)

Now we can plug these two endpoints into the distance formula:

$\small \sqrt{(x_{1}- x_{2})^2 + (y_{1} - y_{2})^2 }$ note that it really does not matter which pair we use as $\small (x_{1}, y_{1})$ and which as $\small (x_{2}, y_{2})$ since we'll be squaring these differences anyway, just as long as we are consistent.

$\small \sqrt{(28 - 10)^2 + (8 - - 1 )^2 }$

$\small \sqrt{18^2 + 9^2 }$

$\small \sqrt{324 + 81 } = \sqrt{405}=9\sqrt{5}$

Report an Error

Example Question #1 : How To Find The Length Of A Line With Distance Formula

Find the length of $y = \frac{1}{5} x + 7$ for the interval $-5 \leq x \leq 10$ .

Possible Answers:

5.831

16.155

15.033

15.297

5.099

Correct answer:

15.297

Explanation:

To find this length, we need to know the y-coordinates for the endpoints.

First, plug in -5 for x:

$y = \frac{1}{5}(-5) + 7 = -1 + 7 = 6$

Next, plug in 10 for x:

$y = \frac{1}{5} (10 ) + 7 = 2 + 7 = 9$

So we are finding the distance between the points (-5, 6) and (10, 9 )

We will use the distance formula, $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2 ) ^2 }$ . We could assign either point as (x_1, y_1) and it would still work, but let's choose (10, 9 ) :

$\sqrt{(10 - - 5 )^2 + (9 - 6 )^ 2 } = \sqrt{15^2 + 3^2 } = \sqrt{225 + 9 } = \sqrt{234} \approx 15.297$

Report an Error

Example Question #11 : Distance Formula

Find the length of y = -2x + 3 for the interval $-10 \leq y \leq 15$ .

Possible Answers:

13.463

22.355

25.005

9.657

27.951

Correct answer:

27.951

Explanation:

First, we need to figure out the x-coordinates of the endpoints so that we can use the distance formula, $\sqrt{(x_1 - x_2 )^2 + (y_1 - y_2 ) ^ 2 }$

Plug in -10 for y and solve for x:

-10 = -2x + 3 subtract 3 from both sides

-13 = -2x divide both sides by -2

6.5 = x

Plug in 15 for y and solve for x:

15 = -2x + 3 subtract 3 from both sides

12 = -2x divide both sides by -2

-6 = x

The endpoints are (6.5, -10) and (-6, 15 ) . We could choose either point to be (x_1, y_1) . Let's choose (6.5, -10 ) .

$\sqrt{(6.5 -- 6 )^2 + (-10 - 15 )^2 }= \sqrt { 12.5^2 + (-25)^2 } = \sqrt{156.25 + 625}= \sqrt{781.25} \approx 27.951$

Report an Error

Example Question #12 : Distance Formula

Find the length of the line 2x - 4 y + 10 = 0 for the interval $-5 \leq x \leq 0$ .

Possible Answers:

7.500

4.330

5.590

2.500

15.811

Correct answer:

5.590

Explanation:

To calculate the distance, first find the y-coordinates of the endpoints by plugging the x-coordinates into the equation.

First plug in -5

2(-5) - 4y + 10 = 0

-10 -4y + 10 = 0 combining like terms, we get -10 + 10 is 0

-4 y = 0 divide by -4

y = 0

Now plug in 0

2(0 ) - 4 y + 10 = 0

-4y + 10 = 0 subtract 10 from both sides

-4y = -10 divide by -4

y = 2.5

The endpoints are (-5, 0 ) and (0, 2.5) , and now we can plug these points into the distance formula:

$\sqrt{(-5 - 0 )^2 + (0 - 2.5)^2 } = \sqrt{(-5)^2 + (-2.5)^2 } = \sqrt{25 + 6.25} \approx 5.590$

Report an Error

Example Question #13 : Distance Formula

Find the length of $y = \frac{1}{4}x + 3$ on the interval $- 8 \leq x \leq 12$ .

Possible Answers:

20.616

8.602

21.190

6.403

8.062

Correct answer:

20.616

Explanation:

To find the length, we need to first find the y-coordinates of the endpoints.

First, plug in -8 for x:

$y = \frac{1}{4} (-8) + 3 = -2 + 3 = 1$

Now plug in 12 for x:

$y = \frac{1}{4} (12 ) + 3 = 3 + 3 = 6$

Our endpoints are (-8, 1) and (12, 6 ) .

To find the length, plug these points into the distance formula:

$\sqrt{(6-1)^2 + (12--8)^2 } = \sqrt{5^2 + 20^2 } = \sqrt{25 + 400 } \approx 20.616$

Report an Error

Example Question #14 : Distance Formula

Jose is walking from his house to the grocery store. He walks 120 feet north, then turns left to walk another 50 feet west. On the way back home, Jose finds a straight line shortcut back to his house. How long is this shortcut?

Possible Answers:

$150\ ft.$

$130\ ft.$

$170\ ft.$

$90\ ft.$

Correct answer:

$130\ ft.$

Explanation:

When walking north and then taking a left west, a 90 degree angle is formed. When Jose returns home going in a straight line, this will now form the hypotenuse of a right triangle. The legs of the triangle are 120 ft and 50 ft respectively.

To solve, use the pythagorean formula.

$50^{2}+120^{2}=c^{2}$

$2500+14400=c^{2}$

$16900=c^{2}$

$\sqrt{16900}=\sqrt{c^{2}}$

130=c

130 ft is the straight line distance home.

The distance formula could also be used to solve this problem.

We will assume that home is at the point (0,0)

$Distance=\sqrt{(50-0)^{2}+(120-0)^{2}}$

Distance = 130 ft.

Report an Error

Example Question #11 : How To Find The Length Of A Line With Distance Formula

A line has endpoints at (8,4) and (5,10). How long is this line?

Possible Answers:

$5\sqrt{3}$

$\sqrt{5}$

None of these.

$3\sqrt{5}$

Correct answer:

$3\sqrt{5}$

Explanation:

We find the exact length of lines using their endpoints and the distance formula.

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Given the endpoints,

(x_1,y_1)=(8,4)

(x_2,y_2)=(5, 10)

the distance formula becomes,

$\sqrt{(5-8)^2+(10-4)^2}=\sqrt{9+36}=\sqrt{45}=\sqrt{9\cdot 5}=\mathbf{3\sqrt{5}}$ .

Report an Error

← Previous 1 2 3 4 5 6 7 8 9 10 11 … 22 23 Next →

View Tutors

Steven
Certified Tutor

University of Nevada-Las Vegas, Bachelor of Science, Elementary Ed. University of Nevada-Las Vegas, Masters, Curriculum and I...

View Tutors

Abhisek
Certified Tutor

University of Florida, Bachelor of Engineering, Computer Science.

View Tutors

Hannah
Certified Tutor

University of North Florida, Bachelor in Arts, Education.

All Intermediate Geometry Resources

8 Diagnostic Tests 250 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Physics Tutors in Los Angeles, LSAT Tutors in San Diego, LSAT Tutors in Washington DC, Spanish Tutors in Boston, SSAT Tutors in Chicago, SSAT Tutors in Atlanta, ACT Tutors in Philadelphia, Math Tutors in Los Angeles, English Tutors in Phoenix, Physics Tutors in Philadelphia

Popular Courses & Classes

GMAT Courses & Classes in Denver, LSAT Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Washington DC, ACT Courses & Classes in Chicago, MCAT Courses & Classes in Houston, LSAT Courses & Classes in Seattle, Spanish Courses & Classes in Atlanta, MCAT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Atlanta, Spanish Courses & Classes in Los Angeles

Popular Test Prep

SSAT Test Prep in Washington DC, GMAT Test Prep in Chicago, LSAT Test Prep in Seattle, ISEE Test Prep in Chicago, SAT Test Prep in New York City, SSAT Test Prep in New York City, SSAT Test Prep in Miami, SSAT Test Prep in Phoenix, LSAT Test Prep in San Diego, ISEE Test Prep in Phoenix

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

Intermediate Geometry : Lines

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #6 : How To Find The Length Of A Line With Distance Formula

Example Question #7 : How To Find The Length Of A Line With Distance Formula

Example Question #8 : How To Find The Length Of A Line With Distance Formula

Example Question #9 : How To Find The Length Of A Line With Distance Formula

Example Question #1 : How To Find The Length Of A Line With Distance Formula

Example Question #11 : Distance Formula

Example Question #12 : Distance Formula

Example Question #13 : Distance Formula

Example Question #14 : Distance Formula

Example Question #11 : How To Find The Length Of A Line With Distance Formula

All Intermediate Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: