Midpoint Formula - SSAT Upper Level Quantitative (Math)

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Question

A line segment on the coordinate plane has one endpoint at $\left ( 8, 10 \right )$ ; its midpoint is $\left ( A+B, A- B\right )$ . Which of the following gives the -coordinate of its other endpoint in terms of and ?

Answer

To find the value of the -coordinate of the other endpoint, we will assign the variable . Then, since the -coordinate of the midpoint of the segment is the mean of those of its endpoints, the equation that we can set up is

$\frac{T + 8 }{2} = A + B$ .

We solve for :

$\frac{T + 8 }{2} \cdot 2 =\left ( A + B \right )\cdot 2$

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Question

A line segment on the coordinate plane has midpoint $\left ( A + 5, B + 7\right )$ . One of its endpoints is $\left ( 10, 6 \right )$ . What is the -coordinate of the other endpoint, in terms of and/or ?

Answer

Let be the -coordinate of the other endpoint. Since the -coordinate of the midpoint of the segment is the mean of those of the endpoints, we can set up an equation as follows:

$\frac{Y + 6 }{2} = B + 7$

$\frac{Y + 6 }{2} \cdot 2= \left ( B + 7 \right )\cdot 2$

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Question

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 . Give the -coordinate of the other endpoint of the segment.

Answer

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 .

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Question

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.

Answer

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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Question

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

Give the -coordinate of .

Answer

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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Question

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

Give the -coordinate of .

Answer

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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Question

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 .

Answer

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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Question

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 .

Answer

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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Question

A line segment on the coordinate plane has endpoints and . In terms of and , as applicable, give the -coordinate of its midpoint.

Answer

The -coordinate of the midpoint of a line segment is the mean of the -coordinates of its endpoints. Therefore, the -coordinate is

$y= \frac{(a-4)+(-b)}{2} = \frac{a-b-4}{2}$ .

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Question

A line segment on the coordinate plane has endpoints and . In terms of and , as applicable, give the -coordinate of its midpoint.

Answer

The -coordinate of the midpoint of a line segment is the mean of the -coordinates of its endpoints. Therefore, the -coordinate is

$x= \frac{(a-5)+(b+8)}{2} = \frac{a+b-5+8}{2} = \frac{a+b+3}{2}$ .

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Question

A line segment has the endpoints and . What is the midpoint of the line?

Answer

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

$\text{Midpoint}=(\frac{6+(-2)}{2}, \frac{10+(-4)}{2})=(\frac{4}{2}, \frac{6}{2})=(2,3)$

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Question

What is the midpoint of a line segment with endpoints and ?

Answer

To find the midpoint of a line, just take the average of the -coordinates and the average of the -coordinates.

$\text{Midpoint}=(\frac{12+(-2)}{2}, \frac{2+1}{2})=(\frac{10}{2}, \frac{3}{2})=(5, \frac{3}{2})$

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Question

What is the midpoint of a line segments with endpoints and ?

Answer

To find the midpoint of a line segment, take the average of the -coordinates, then take the average of the -coordinates.

$\text{Midpoint}=(\frac{2a+4a}{2}, \frac{-b+5b}{2})$

$\text{Midpoint}=(\frac{6a}{2}, \frac{4b}{2})=(3a, 2b)$

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Question

A line has the endpoints $(8,8) \text{ and }(2, -1)$ . What is its midpoint?

Answer

The coordinates of the midpoint of a line are just the averages of the coordinates of the endpoints.

$\text{Midpoint}=\left(\frac{8+2}{2}, \frac{8-1}{2}\right)=\left(5, \frac{7}{2}\right)$

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Question

A line segment has endpoints $(-2, 0) \text{ and }(1, 5)$ . What is the midpoint of this line segment?

Answer

To find the coordinates of the midpoint of a line, take the averages of the x and y coordinates of the endpoints.

$\text{Midpoint}=\left(\frac{-2+1}{2}, \frac{5+0}{2}\right)=\left(-\frac{1}{2}, \frac{5}{2}\right)$

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Question

What is the midpoint of a line if the endpoints of the line are: $(6,9) \textup{ and }(3,-2)$ ?

Answer

Write the midpoint formula.

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

Substitute the values.

$M=\left(\frac{6+3}{2},\frac{9-2}{2}\right)= \left(\frac{9}{2},\frac{7}{2}\right)$

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