Midpoint Formula - SSAT Upper Level Quantitative

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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 ?

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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 ?

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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.

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Answer

Using the part of the midpoint formula

.

set and solve:

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

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.

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Answer

In the part of the midpoint formula

,

set , and solve:

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

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

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Answer

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

,

set , and solve:

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

Now, find the -coordinate of similarly, setting

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

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 .

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Answer

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

,

set , and solve:

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

Now, find the -coordinate of similarly, setting

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

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 .

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Answer

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

,

set , and solve:

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

Now, find the -coordinate of similarly, setting

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

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 .

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Answer

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

,

set , and solve:

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

Now, find the -coordinate of similarly, setting

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

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.

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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.

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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?

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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 ?

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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 ?

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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?

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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?

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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: $(6,9) \textup{ and }(3,-2)$ ?

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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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Question

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

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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.

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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?

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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 ?

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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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