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SSAT Upper Level Math : Midpoint Formula

Study concepts, example questions & explanations for SSAT Upper Level Math

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All SSAT Upper Level Math Resources

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Example Questions

Example Question #3 : How To Find The Endpoints Of A Line Segment

One endpoint of a line segment on the coordinate plane is the point $\left (2\frac{4}{5}, -1\frac{1}{5} \right )$ ; the midpoint of the segment is the point $\left (6\frac{1}{2}, 4\frac{1}{2} \right )$ . Give the -coordinate of the other endpoint of the segment.

Possible Answers:

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

$4 \frac{3}{20}$

$10\frac{1}{5}$

$3\frac{7}{10}$

$9\frac{3}{10}$

Correct answer:

$10\frac{1}{5}$

Explanation:

In the part of the midpoint formula

$x_{M}= \frac{x_{1} + x_{2} }{2}$ ,

set $x_{M}= 6\frac{1}{2}, x_{2}= 2\frac{4}{5}$ , and solve:

$x_{M}= \frac{x_{1} + x_{2} }{2}$

$6\frac{1}{2}= \frac{x_{1} + 2\frac{4}{5} }{2}$

$6\frac{1}{2} \cdot 2 = x_{1} + 2\frac{4}{5}$

$x_{1} + 2\frac{4}{5} = 13$

$x_{1} = 10\frac{1}{5}$

This is the correct -coordinate.

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Example Question #4 : How To Find The Endpoints Of A Line Segment

One endpoint of a line segment on the coordinate plane is the point $\left (8.32, -4.29 \right )$ ; the midpoint of the segment is the point (-1.89, 8.56) . Give the -coordinate of the other endpoint of the segment.

Possible Answers:

-12.1

3.215

2.135

21.41

-12.85

Correct answer:

21.41

Explanation:

Using the part of the midpoint formula

$y_{M}= \frac{y_{1} + y_{2} }{2}$ .

set $y_{M}=8.56, y_{2}= -4.29$ and solve:

$8.56= \frac{y_{1} -4.29 }{2}$

$y_{1} -4.29 = 17.12$

$y_{1} = 21.41$

The second endpoint is ( -12.1, 21.41) .

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Example Question #4 : How To Find The Endpoints Of A Line Segment

On the coordinate plane, is the midpoint of $\overline{AB}$ and is the midpoint of $\overline{AM}$ . has coordinates (288, 108) and has coordinates $\left ( 176, 300\right )$ .

Give the -coordinate of .

Possible Answers:

102

876

156

Correct answer:

876

Explanation:

First, find the -coordinate of . In the part of the midpoint formula

$y_{M}= \frac{y_{1} + y_{2} }{2}$ ,

set $y_{M}=300, y_{2}= 108$ , and solve:

$300= \frac{y_{1} + 108}{2}$

$y_{1} + 108 = 600$

$y_{1} =492$

Now, find the -coordinate of similarly, setting

$y_{M}= 492,y_{2}= 108$

$492= \frac{y_{1} + 108}{2}$

$y_{1} + 108= 984$

$y_{1} = 876$

This is the correct response.

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Example Question #6 : How To Find The Endpoints Of A Line Segment

On the coordinate plane, is the midpoint of $\overline{AB}$ and is the midpoint of $\overline{AM}$ . has coordinates $\left (4\frac{1}{3},-12\frac{1}{2} \right )$ and has coordinates $\left ( -7 \frac{5}{6}, 6\frac{1}{3}\right )$ .

Give the -coordinate of .

Possible Answers:

$40\frac{5}{6}$

$62\frac{5}{6}$

$-44\frac{1}{3}$

$1\frac{1}{4}$

Correct answer:

$-44\frac{1}{3}$

Explanation:

First, find the -coordinate of . In the part of the midpoint formula

$x_{M}= \frac{x_{1} + x_{2} }{2}$ ,

set $x_{M}= -7 \frac{5}{6}, x_{2}= 4\frac{1}{3}$ , and solve:

$-7 \frac{5}{6}= \frac{x_{1} + 4\frac{1}{3}}{2}$

$x_{1} + 4\frac{1}{3}= -7 \frac{5}{6} \cdot 2$

$x_{1} + 4\frac{1}{3}= -15 \frac{2}{3}$

$x_{1} = -15 \frac{2}{3} - 4\frac{1}{3}$

$x_{1} = -20$

Now, find the -coordinate of similarly, setting

$x_{M}= -20, x_{2}= 4\frac{1}{3}$

$-20 = \frac{x_{1} + 4\frac{1}{3}}{2}$

$x_{1} + 4\frac{1}{3}= -20 \cdot 2$

$x_{1} + 4\frac{1}{3}= -40$

$x_{1} = -40 - 4\frac{1}{3}$

$x_{1} = -44\frac{1}{3}$

This is the correct response.

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

Give the -coordinate of .

Possible Answers:

$-44\frac{1}{3}$

$62\frac{5}{6}$

$-50\frac{1}{6}$

$37\frac{5}{6}$

$-24\frac{2}{3}$

Correct answer:

$62\frac{5}{6}$

Explanation:

First, find the -coordinate of . In the part of the midpoint formula

$y_{M}= \frac{y_{1} + y_{2} }{2}$ ,

set $y_{M}=6\frac{1}{3}, y_{2}= -12\frac{1}{2}$ , and solve:

$6\frac{1}{3}= \frac{y_{1} -12\frac{1}{2} }{2}$

$y_{1} -12\frac{1}{2}= 6\frac{1}{3} \cdot 2$

$y_{1} -12\frac{1}{2}=12 \frac{2}{3}$

$y_{1} =12 \frac{2}{3} + 12\frac{1}{2}$

$y_{1} = 25\frac{1}{6}$

Now, find the -coordinate of similarly, setting

$y_{M}=25\frac{1}{6}, y_{2}= -12\frac{1}{2}$

$25\frac{1}{6}= \frac{y_{1} -12\frac{1}{2} }{2}$

$y_{1} -12\frac{1}{2} = 25\frac{1}{6} \cdot 2$

$y_{1} -12\frac{1}{2} = 50\frac{1}{3}$

$y_{1} = 50\frac{1}{3} +12\frac{1}{2}$

$y_{1} = 62\frac{5}{6}$

This is the correct response.

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Example Question #3 : How To Find The Endpoints Of A Line Segment

On the coordinate plane, is the midpoint of $\overline{AB}$ and is the midpoint of $\overline{AM}$ . has coordinates (288, 108) and has coordinates $\left ( 176, 300\right )$ .

Give the -coordinate of .

Possible Answers:

684

736

204

116

Correct answer:

Explanation:

First, find the -coordinate of . In the part of the midpoint formula using the coordinates from and

$x_{M}= \frac{x_{1} + x_{2} }{2}$ ,

set $x_{M}= 176, x_{2}= 288$ , and solve:

$176= \frac{x_{1} + 288}{2}$

$x_{1} + 288 = 352$

$x_{1} =64$

Now, find the -coordinate of similarly, setting

$x_{M}= 64, x_{2}= 288$

$64= \frac{x_{1} + 288}{2}$

$x_{1} + 288 = 128$

$x_{1} = -160$

This is the correct response.

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SSAT Upper Level Math : Midpoint Formula

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #3 : How To Find The Endpoints Of A Line Segment

Example Question #4 : How To Find The Endpoints Of A Line Segment

Example Question #4 : How To Find The Endpoints Of A Line Segment

Example Question #6 : How To Find The Endpoints Of A Line Segment

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

Example Question #3 : How To Find The Endpoints Of A Line Segment

All SSAT Upper Level Math Resources

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