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Algebra 1 : Midpoint Formula

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Example Questions

Example Question #861 : Functions And Lines

Find the midpoint of the line segment that has endpoints at $(1, 3)\text{ and }(9, 5)$ .

Possible Answers:

(5, 4)

$\left(\frac{9}{2}, \frac{15}{2}\right)$

(2, 7)

(4, 5)

Correct answer:

(5, 4)

Explanation:

To find the midpoint of a line segment, take the average of the x-coordinates and the average of the y-coordinates.

$\text{Midpoint}=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

$\text{Midpoint}=\left(\frac{1+9}{2}, \frac{3+5}{2}\right)=(5, 4)$

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Example Question #862 : Functions And Lines

Find the midpoint of the line segment that has endpoints at $(-2, 6)\text{ and }(-8, 10)$ .

Possible Answers:

(-5, 8)

(8, 30)

(2, 1)

(8, 5)

Correct answer:

(-5, 8)

Explanation:

To find the midpoint of a line segment, take the average of the x-coordinates and the average of the y-coordinates.

$\text{Midpoint}=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

$\text{Midpoint}=\left(\frac{-2+(-8)}{2}, \frac{6+10}{2}\right)=(-5, 8)$

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Example Question #33 : How To Find The Midpoint Of A Line Segment

Find the midpoint of a line segment that has endpoints at $\left(\frac{1}{2}, 6\right)$ and $\left(-\frac{7}{2}, 18\right)$ .

Possible Answers:

$\left(-\frac{5}{2}, 4\right)$

$\left(-\frac{3}{2}, 6\right)$

(-2, 6)

$\left(-\frac{3}{2}, 12\right)$

Correct answer:

$\left(-\frac{3}{2}, 12\right)$

Explanation:

To find the midpoint of a line segment, take the average of the x-coordinates and the average of the y-coordinates.

$\text{Midpoint}=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

$\text{Midpoint}=\left(\frac{\frac{1}{2}+(-\frac{7}{2})}{2}, \frac{6+18}{2}\right)=\left(-\frac{3}{2},12\right)$

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Example Question #871 : Functions And Lines

Find the midpoint of the line segment that has endpoints at $(7, 0) \text{ and }(8,9)$ .

Possible Answers:

(15, 9)

$\left(\frac{7}{2}, \frac{17}{2}\right)$

$\left(\frac{1}{2}, -\frac{9}{2}\right)$

$\left(\frac{15}{2}, \frac{9}{2}\right)$

Correct answer:

$\left(\frac{15}{2}, \frac{9}{2}\right)$

Explanation:

To find the midpoint of a line segment, take the average of the x-coordinates and the average of the y-coordinates.

$\text{Midpoint}=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

$\text{Midpoint}=\left(\frac{7+8}{2}, \frac{0+9}{2}\right)=\left(\frac{15}{2}, \frac{9}{2}\right)$

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Example Question #872 : Functions And Lines

Find the midpoint of the line segment that has endpoints at $(100, 5)\text{ and }(2, 0)$ .

Possible Answers:

$\left(50, \frac{7}{2}\right)$

$\left(51, \frac{5}{2}\right)$

(50, 1)

(100, 0)

Correct answer:

$\left(51, \frac{5}{2}\right)$

Explanation:

To find the midpoint of a line segment, take the average of the x-coordinates and the average of the y-coordinates.

$\text{Midpoint}=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

$\text{Midpoint}=\left(\frac{100+2}{2}, \frac{5+0}{2}\right)=\left(51, \frac{5}{2}\right)$

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Example Question #873 : Functions And Lines

Find the midpoint of the line segment that has endpoints at $(-5, -2)\text{ and }(-9, -10)$ .

Possible Answers:

(-7, -6)

(7, 2)

(-2, -4)

$\left(-\frac{15}{2}, -\frac{11}{2}\right)$

Correct answer:

(-7, -6)

Explanation:

To find the midpoint of a line segment, take the average of the x-coordinates and the average of the y-coordinates.

$\text{Midpoint}=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

$\text{Midpoint}=\left(\frac{-5+(-9)}{2}, \frac{-2+(-10)}{2}\right)=(-7, -6)$

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Example Question #874 : Functions And Lines

Find the midpoint of the line segment that has endpoints at $(1, -2)\text{ and }(7, -12)$ .

Possible Answers:

(4, -7)

(-4, -7)

(3, 5)

$\left(-\frac{13}{2}, \frac{9}{2}\right)$

Correct answer:

(4, -7)

Explanation:

To find the midpoint of a line segment, take the average of the x-coordinates and the average of the y-coordinates.

$\text{Midpoint}=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

$\text{Midpoint}=\left(\frac{1+7}{2}, \frac{-2+(-12)}{2}\right)=(4, -7)$

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Example Question #875 : Functions And Lines

Find the midpoint of the line segment that has endpoints at $(-0.9, 1)\text{ and }(0.1, 5)$ .

Possible Answers:

(-0.4, 3)

(-0.5, 3)

(-0.5, -0.5)

(-0.5, 2)

Correct answer:

(-0.4, 3)

Explanation:

To find the midpoint of a line segment, take the average of the x-coordinates and the average of the y-coordinates.

$\text{Midpoint}=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

$\text{Midpoint}=\left(\frac{-0.9+0.1}{2}, \frac{1+5}{2}\right)=(-0.4, 3)$

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Example Question #876 : Functions And Lines

Find the midpoint of a line segment that has endpoints at $(11, -5)\text{ and }(1, -9)$ .

Possible Answers:

(6, -2)

(2, 1)

(5, 4)

(6, -7)

Correct answer:

(6, -7)

Explanation:

To find the midpoint of a line segment, take the average of the x-coordinates and the average of the y-coordinates.

$\text{Midpoint}=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

$\text{Midpoint}=\left(\frac{11+1}{2}, \frac{-5+(-9)}{2}\right)=(6, -7)$

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Example Question #877 : Functions And Lines

Find the midpoint of a line segment that has endpoints at $(-7, 12)\text{ and }(-3, -2)$ .

Possible Answers:

(-2, -2)

(5, -5)

(-5, 5)

(-2, 6)

Correct answer:

(-5, 5)

Explanation:

To find the midpoint of a line segment, take the average of the x-coordinates and the average of the y-coordinates.

$\text{Midpoint}=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

$\text{Midpoint}=\left(\frac{-7+(-3)}{2}, \frac{12+(-2)}{2}\right)=(-5, 5)$

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Algebra 1 : Midpoint Formula

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #861 : Functions And Lines

Example Question #862 : Functions And Lines

Example Question #33 : How To Find The Midpoint Of A Line Segment

Example Question #871 : Functions And Lines

Example Question #872 : Functions And Lines

Example Question #873 : Functions And Lines

Example Question #874 : Functions And Lines

Example Question #875 : Functions And Lines

Example Question #876 : Functions And Lines

Example Question #877 : Functions And Lines

All Algebra 1 Resources

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