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High School Math : Algebra I

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8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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Example Question #11 :Midpoint And Distance Formulas

What is the midpoint of a line segment whose endpoints are $(-1,-7)\; and\; (3,11)$ ?

Possible Answers:

(1,5)

(2,-1)

(1,2)

(2,9)

(0,0)

Correct answer:

(1,2)

Explanation:

The midpoint is the average of the endpoints.

$(\frac{3-1}{2}, \frac{11-7}{2})=(\frac{2}{2},\frac{4}{2})=(1,2)$

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Example Question #11 :How To Find The Midpoint Of A Line Segment

What is the midpoint of a line segment connecting the points (-4,6) and (8,2) ?

Possible Answers:

(-2,4)

(-6,2)

(2,-6)

(2,4)

(4,2)

Correct answer:

(2,4)

Explanation:

Use the midpoint formula, $mid=(\frac{x_{1}+x_2}{2}, \frac{y_1+y_2}{2})$ , with our points (-4,6) and (8,2) .

$mid=(\frac{-4+8}{2},\frac{6+2}{2})$

$mid=(\frac{4}{2},\frac{8}{2})=(2,4)$

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Example Question #13 :Midpoint And Distance Formulas

Find the midpoint between these two points:

(4,-1) and (-12,3)

Possible Answers:

(4,-1)

(-4,1)

(-8,2)

(8,-2)

(-4,-1)

Correct answer:

(-4,1)

Explanation:

To find the midpoint of a line segment, use the standard midpoint equation:

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Plug in the given points and simplify:

$(\frac{-12+4}{2},\frac{-1+3}{2})=(-4,1)$

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Example Question #12 :Midpoint And Distance Formulas

Find the midpoint of the line segment connecting (-1, 5) and (5, -3) .

Possible Answers:

(4, 2)

(2, 1)

(-6, 8)

(-2, 1)

(4, -2)

Correct answer:

(2, 1)

Explanation:

To find the midpoint of a line segment, we find the average of the-values and the average of the-values. Thus, we have that:

Midpoint = $\left(\frac{-1 + 5}{2}, \frac{5 + (-3)}{2}\right) = \left(\frac{4}{2}, \frac{2}{2}\right) = (2, 1)$

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Example Question #1 :How To Find The Endpoints Of A Line Segment

What is the length of a line with endpoints (0,3) and (5,15) ?

Possible Answers:

$\frac{5}{12}$

Correct answer:

Explanation:

The formula for the length of a line is very similiar to the pythagorean theorem:

(x_2-x_1)^2+(y_2-y_1)^2=l^2

Plug in our given numbers to solve:

(x_2-x_1)^2+(y_2-y_1)^2=l^2

(5-0)^2+(15-3)^2=l^2

5^2+12^2=l^2

25+144=l^2

169=l^2

$\sqrt{169}=\sqrt{l^2}$

13=l

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Example Question #2 :How To Find The Endpoints Of A Line Segment

The points A, B, and C reside on a line segment.Bis the midpoint ofAC.If line AB measures 6 units in length, what is the length of lineAC?

Possible Answers:

Correct answer:

Explanation:

IfBis the midpoint ofAC,thenACis twice as long asAB.We are told thatAB=6.

A------B------C

The diagram shows six units between pointsAandB,withBas the midpoint of segmentAC. Therefore segmentBCis also six units long, so lineACis twelve units long.

$\small (2)(6)=12$

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

What is the length of a line segment with end points (0,1) and (5,7) ?

Possible Answers:

$\sqrt{61}$

7.25

9.13

$\sqrt{68}$

Correct answer:

$\sqrt{61}$

Explanation:

The formula for the length of line is the distance formula, which is very similar to the Pythagorean theorem.

(x_2-x_1)^2+(y_2-y_1)^2=l^2

Plug in the given values and solve for the length.

(5-0)^2+(7-1)^2=l^2

(5)^2+(6)^2=l^2

25+36=l^2

61=l^2

$\sqrt{61}=\sqrt{l^2}$

$\sqrt{61}=l$

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Example Question #2 :How To Find The Length Of A Line With Distance Formula

What would be the length of a line with endpoints at (1,6) and (9,9) ?

Possible Answers:

4.5

$\sqrt{39}$

8.13

$\sqrt{73}$

$\sqrt{91}$

Correct answer:

$\sqrt{73}$

Explanation:

The formula for the length of line is the distance formula, which is very similar to the Pythagorean theorem.

(x_2-x_1)^2+(y_2-y_1)^2=l^2

Plug in the given values and solve for the length.

(9-1)^2+(9-6)^2=l^2

(8)^2+(3)^2=l^2

64+9=l^2

73=l^2

$\sqrt{73}=\sqrt{l^2}$

$\sqrt{73}=l$

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Example Question #3 :How To Find The Length Of A Line With Distance Formula

What line goes through the points (-1,5) and (5,15) ?

Possible Answers:

-4x-3y=9

-2x+5y=10

-5x+3y=20

3x+2y=5

3x-5y=15

Correct answer:

-5x+3y=20

Explanation:

Find the slope between the two points:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{15-5}{5-(-1)}=\frac{10}{6}=\frac{5}{3}$

Next, use the slope-intercept form of the equation:

y=mx+b or $15=\frac{5}{3}(5)+b$ where $b=\frac{20}{3}$

So the equation becomes $y=\frac{5}{3}x+ \frac{20}{3}$ or in standard form -5x+3y=20 .

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

What is the length of a line with endpoints at (4,10) and (8,13) ?

Possible Answers:

625

Correct answer:

Explanation:

The formula for the length of a line is very similiar to the pythagorean theorem:

(x_2-x_1)^2+(y_2-y_1)^2=l^2

Plug in our given numbers to solve:

(x_2-x_1)^2+(y_2-y_1)^2=l^2

(8-4)^2+(13-10)^2=l^2

4^2+3^2=l^2

16+9=l^2

25=l^2

$\sqrt{25}=\sqrt{l^2}$

5=l

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High School Math : Algebra I

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #11 :Midpoint And Distance Formulas

Example Question #11 :How To Find The Midpoint Of A Line Segment

Example Question #13 :Midpoint And Distance Formulas

Example Question #12 :Midpoint And Distance Formulas

Example Question #1 :How To Find The Endpoints Of A Line Segment

Example Question #2 :How To Find The Endpoints Of A Line Segment

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

Example Question #2 :How To Find The Length Of A Line With Distance Formula

Example Question #3 :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

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