How to find the endpoints of a line segment - SSAT Upper Level Quantitative (Math)

Card 0 of 8

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 :

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

Compare your answer with the correct one above

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:

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

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

Compare your answer with the correct one above

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 .

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

Tap the card to reveal the answer