Coordinate Geometry

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SSAT Middle Level Quantitative › Coordinate Geometry

Questions 1 - 10

1

Give the slope of a line that passes through and .

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{3}{4}$

$\frac{2}{5}$

$\frac{4}{5}$

Explanation

Using the slope formula, substituting $x_{1}=5$ , $x_{2}=2$ , $y_{1}=5$ , and $y_{2}=4$ :

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{4-5}{2-5}$

Subtract to get:

$m=\frac{-1}{-3}$

Cancel out the negative signs to get:

$m=\frac{1}{3}$

2

Give the slope of a line that passes through and .

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{3}{4}$

$\frac{2}{5}$

$\frac{4}{5}$

Explanation

Using the slope formula, substituting $x_{1}=5$ , $x_{2}=2$ , $y_{1}=5$ , and $y_{2}=4$ :

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{4-5}{2-5}$

Subtract to get:

$m=\frac{-1}{-3}$

Cancel out the negative signs to get:

$m=\frac{1}{3}$

3

A line segment on the coordinate plane has its endpoints at the points with coordinates $\left ( 4 \frac{1}{3}, - 4 \frac{1}{3}\right )$ and $\left ( 12, 12\right )$ . Give the coordinates of the midpoint of the segment.

$\left ( 8 \frac{1}{6}, 3 \frac{5}{6} \right )$

$\left ( 16 \frac{1}{3}, 7 \frac{2}{3} \right )$

$\left ( 7 \frac{2}{3}, 16 \frac{1}{3}\right )$

$\left ( 3 \frac{5}{6} , 8 \frac{1}{6} \right )$

Explanation

The -coordinate of the midpoint is half the sum of the -coordinates of the endpoints:

$\frac{1}{2} \left ( 4 \frac{1}{3} + 12 \right )$

$= \frac{1}{2} \times 16 \frac{1}{3}$

$= \frac{1}{2} \times \frac{49}{3}$

$= \frac{49}{6}$

$=8 \frac{1}{6}$

The -coordinate of the midpoint is found similarly:

$\frac{1}{2} \left (- 4 \frac{1}{3} + 12 \right )$

$= \frac{1}{2} \left (12 - 4 \frac{1}{3} \right )$

$= \frac{1}{2} \times 7 \frac{2}{3}$

$= \frac{1}{2} \times \frac{23}{3}$

$= \frac{23}{6}$

$= 3 \frac{5}{6}$

The midpoint has coordinates $\left ( 8 \frac{1}{6}, 3 \frac{5}{6} \right )$ .

4

A line segment on the coordinate plane has its endpoints at the points with coordinates $\left ( 4 \frac{1}{3}, - 4 \frac{1}{3}\right )$ and $\left ( 12, 12\right )$ . Give the coordinates of the midpoint of the segment.

$\left ( 8 \frac{1}{6}, 3 \frac{5}{6} \right )$

$\left ( 16 \frac{1}{3}, 7 \frac{2}{3} \right )$

$\left ( 7 \frac{2}{3}, 16 \frac{1}{3}\right )$

$\left ( 3 \frac{5}{6} , 8 \frac{1}{6} \right )$

Explanation

The -coordinate of the midpoint is half the sum of the -coordinates of the endpoints:

$\frac{1}{2} \left ( 4 \frac{1}{3} + 12 \right )$

$= \frac{1}{2} \times 16 \frac{1}{3}$

$= \frac{1}{2} \times \frac{49}{3}$

$= \frac{49}{6}$

$=8 \frac{1}{6}$

The -coordinate of the midpoint is found similarly:

$\frac{1}{2} \left (- 4 \frac{1}{3} + 12 \right )$

$= \frac{1}{2} \left (12 - 4 \frac{1}{3} \right )$

$= \frac{1}{2} \times 7 \frac{2}{3}$

$= \frac{1}{2} \times \frac{23}{3}$

$= \frac{23}{6}$

$= 3 \frac{5}{6}$

The midpoint has coordinates $\left ( 8 \frac{1}{6}, 3 \frac{5}{6} \right )$ .

5

Give the slope of the line that passes through and .

$\frac{1}{2}$

$\frac{3}{2}$

$-\frac{1}{2}$

$-\frac{3}{4}$

Explanation

Using the slope formula for

$x_{1}=-5$ , $x_{2}=1$ , $y_{1}=3$ , and $y_{2}=9$ :

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{9-3}{1-(-5)}$

Combine the negative signs to get:

$m=\frac{9-3}{1+5}$

Subtract and add to get:

$m=\frac{6}{6}$

Reduce to get:

6

Give the slope of a line that passes through and .

$\frac{1}{2}$

$-\frac{1}{2}$

$\frac{3}{7}$

Explanation

Using the slope formula, where is the slope, $(x_{1},y_{1})=$ , and $(x_{2},y_{2})=$ :

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{3-7}{2-4}$

$m=\frac{-4}{-2}$

7

Give the slope of a line that passes through and .

$\frac{1}{2}$

$-\frac{1}{2}$

$\frac{3}{7}$

Explanation

Using the slope formula, where is the slope, $(x_{1},y_{1})=$ , and $(x_{2},y_{2})=$ :

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{3-7}{2-4}$

$m=\frac{-4}{-2}$

8

Give the slope of the line that passes through and .

$\frac{1}{2}$

$\frac{3}{2}$

$-\frac{1}{2}$

$-\frac{3}{4}$

Explanation

Using the slope formula for

$x_{1}=-5$ , $x_{2}=1$ , $y_{1}=3$ , and $y_{2}=9$ :

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{9-3}{1-(-5)}$

Combine the negative signs to get:

$m=\frac{9-3}{1+5}$

Subtract and add to get:

$m=\frac{6}{6}$

Reduce to get:

9

Give the slope of a line that passes through and .

$\frac{1}{9}$

$\frac{2}{9}$

$\frac{4}{7}$

$\frac{3}{8}$

$\frac{5}{7}$

Explanation

Using the slope formula with $x_{1}=-7$ , $x_{2}=2$ , $y_{1}=4$ , $y_{2}=5$ :

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\frac{5-4}{2-(-7)}$

Subtract to get:

$m=\frac{1}{9}$

10

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Identify the correct set of coordinates for rectangle $\small ABCD$ .

$\small (-11,3),(-11,7),(3,3),(3,7)$

$\small (-11,3),(11,7),(3,3),(-3,7)$

$\small (11,3),(-10,7),(3,3),(3,7)$

$\small (-11,3),(-11,7),(-3,-3),(3,7)$

Explanation

To identify the correct set of coordinate points for rectangle $\small ABCD$ , note that the correct set must have two sets of matching $\small x$ coordinates and two sets of matching $\small y$ coordinates. The only answer choice that meets these specifications is: $\small \small (-11,3),(-11,7),(3,3),(3,7)$

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