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SSAT Upper Level Math : Geometry

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All SSAT Upper Level Math Resources

6 Diagnostic Tests 311 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : How To Find The Midpoint Of A Line Segment

A line segment on the coordinate plane has endpoints $\displaystyle (a-5,-b)$ and $\displaystyle (b+8, a-4)$. In terms of $\displaystyle a$ and $\displaystyle b$, as applicable, give the $\displaystyle y$-coordinate of its midpoint.

Possible Answers:

$\frac{a + b + 3}{2}$ $\displaystyle \frac{a + b + 3}{2}$

$\displaystyle a - b - 4$

$\displaystyle a + b + 3$

$\frac{a + b - 4}{2}$ $\displaystyle \frac{a + b - 4}{2}$

$\frac{a - b - 4}{2}$ $\displaystyle \frac{a - b - 4}{2}$

Correct answer:

$\frac{a - b - 4}{2}$ $\displaystyle \frac{a - b - 4}{2}$

Explanation:

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

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

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Example Question #111 : Coordinate Geometry

A line segment has the endpoints (6,10) $\displaystyle (6,10)$ and $\displaystyle (-2, -4)$. What is the midpoint of the line?

Possible Answers:

(8,14) $\displaystyle (8,14)$

(4,6) $\displaystyle (4,6)$

(2, 3) $\displaystyle (2, 3)$

(4,7) $\displaystyle (4,7)$

Correct answer:

(2, 3) $\displaystyle (2, 3)$

Explanation:

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

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

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Example Question #1 : Midpoint Formula

What is the midpoint of a line segment with endpoints (12,2) $\displaystyle (12,2)$ and (-2, 1) $\displaystyle (-2, 1)$?

Possible Answers:

(14,1) $\displaystyle (14,1)$

$(5, \frac{3}{2})$ $\displaystyle (5, \frac{3}{2})$

(10, 3) $\displaystyle (10, 3)$

$(\frac{13}{2}, 0)$ $\displaystyle (\frac{13}{2}, 0)$

Correct answer:

$(5, \frac{3}{2})$ $\displaystyle (5, \frac{3}{2})$

Explanation:

To find the midpoint of a line, just take the average of the $\displaystyle x$-coordinates and the average of the $\displaystyle y$-coordinates.

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

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Example Question #113 : Coordinate Geometry

What is the midpoint of a line segments with endpoints $\displaystyle (2a, -b)$ and $\displaystyle (4a, 5b)$?

Possible Answers:

$\displaystyle (6a, 4b)$

$\displaystyle (3a, 2b)$

$\displaystyle (-a, -2b)$

(3, 4) $\displaystyle (3, 4)$

Correct answer:

$\displaystyle (3a, 2b)$

Explanation:

To find the midpoint of a line segment, take the average of the $\displaystyle x$-coordinates, then take the average of the $\displaystyle y$-coordinates.

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

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

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Example Question #121 : Coordinate Geometry

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

Possible Answers:

$\left(5, \frac{7}{2}\right)$ $\displaystyle \left(5, \frac{7}{2}\right)$

$\left(\frac{7}{2}, 5\right)$ $\displaystyle \left(\frac{7}{2}, 5\right)$

(10, 7) $\displaystyle (10, 7)$

(5, 7) $\displaystyle (5, 7)$

Correct answer:

$\left(5, \frac{7}{2}\right)$ $\displaystyle \left(5, \frac{7}{2}\right)$

Explanation:

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

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

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Example Question #1 : How To Find The Midpoint Of A Line Segment

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

Possible Answers:

$\left(-\frac{1}{2}, \frac{5}{2}\right)$ $\displaystyle \left(-\frac{1}{2}, \frac{5}{2}\right)$

(-1, 5) $\displaystyle (-1, 5)$

$\left(-\frac{1}{2}, 5\right)$ $\displaystyle \left(-\frac{1}{2}, 5\right)$

$\left(5, -\frac{1}{2}\right)$ $\displaystyle \left(5, -\frac{1}{2}\right)$

Correct answer:

$\left(-\frac{1}{2}, \frac{5}{2}\right)$ $\displaystyle \left(-\frac{1}{2}, \frac{5}{2}\right)$

Explanation:

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

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

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Example Question #8 : How To Find The Midpoint Of A Line Segment

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

Possible Answers:

$\left(-\frac{3}{2},\frac{1}{2}\right)$ $\displaystyle \left(-\frac{3}{2},\frac{1}{2}\right)$

$\left(-\frac{3}{2},-\frac{11}{2}\right)$ $\displaystyle \left(-\frac{3}{2},-\frac{11}{2}\right)$

$\left(\frac{9}{2},\frac{7}{2}\right)$ $\displaystyle \left(\frac{9}{2},\frac{7}{2}\right)$

$\left(\frac{7}{2},\frac{9}{2}\right)$ $\displaystyle \left(\frac{7}{2},\frac{9}{2}\right)$

$\left(\frac{3}{2},\frac{11}{2}\right)$ $\displaystyle \left(\frac{3}{2},\frac{11}{2}\right)$

Correct answer:

$\left(\frac{9}{2},\frac{7}{2}\right)$ $\displaystyle \left(\frac{9}{2},\frac{7}{2}\right)$

Explanation:

Write the midpoint formula.

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

Substitute the values.

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

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Example Question #2 : Midpoint Formula

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

Possible Answers:

B + 8 $\displaystyle B + 8$

2B + 20 $\displaystyle 2B + 20$

B + 1 $\displaystyle B + 1$

2B + 1 $\displaystyle 2B + 1$

2B + 8 $\displaystyle 2B + 8$

Correct answer:

2B + 8 $\displaystyle 2B + 8$

Explanation:

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

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

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

$\displaystyle Y + 6 = 2B + 14$

$\displaystyle Y + 6 -6 = 2B + 14 -6$

$\displaystyle Y = 2B + 8$

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Example Question #3 : Midpoint Formula

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

Possible Answers:

$\displaystyle 2 A + 2B+8$

$\displaystyle A + B- 8$

$\displaystyle 2 A + 2B- 8$

$\displaystyle A + B- 16$

A + B+8 $\displaystyle A + B+8$

Correct answer:

$\displaystyle 2 A + 2B- 8$

Explanation:

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

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

We solve for $\displaystyle T$:

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

$\displaystyle T + 8 =2 A + 2B$

$\displaystyle T + 8 - 8 =2 A + 2B- 8$

$\displaystyle T =2 A + 2B- 8$

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Example Question #1 : How To Find The Endpoints Of A Line Segment

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

Possible Answers:

$3\frac{7}{10}$ $\displaystyle 3\frac{7}{10}$

$9\frac{3}{10}$ $\displaystyle 9\frac{3}{10}$

$10\frac{1}{5}$ $\displaystyle 10\frac{1}{5}$

$4 \frac{3}{20}$ $\displaystyle 4 \frac{3}{20}$

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

Correct answer:

$10\frac{1}{5}$ $\displaystyle 10\frac{1}{5}$

Explanation:

In the $\displaystyle x$ part of the midpoint formula

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

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

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

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

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

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

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

This is the correct $\displaystyle x$-coordinate.

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SSAT Upper Level Math : Geometry

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

All SSAT Upper Level Math Resources

Example Questions

Example Question #1 : How To Find The Midpoint Of A Line Segment

Example Question #111 : Coordinate Geometry

Example Question #1 : Midpoint Formula

Example Question #113 : Coordinate Geometry

Example Question #121 : Coordinate Geometry

Example Question #1 : How To Find The Midpoint Of A Line Segment

Example Question #8 : How To Find The Midpoint Of A Line Segment

Example Question #2 : Midpoint Formula

Example Question #3 : Midpoint Formula

Example Question #1 : How To Find The Endpoints Of A Line Segment

All SSAT Upper Level Math Resources

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