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SSAT Upper Level Math : Geometry

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

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All SSAT Upper Level Math Resources

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Example Questions

Example Question #1 : How To Find The Equation Of A Perpendicular Line

Line m passes through the points (1, 4) and (5, 2). If line p is perpendicular to m, then which of the following could represent the equation for p?

Possible Answers:

2x + y = 3

3x + 2y = 4

x – y = 3

2x – y = 3

4x – 3y = 4

Correct answer:

2x – y = 3

Explanation:

The slope of m is equal to ^y2-y1/_x2-x1=^2-4/_5-1= ^-1/₂

Since line p is perpendicular to line m, this means that the products of the slopes of p and m must be –1:

(slope of p) * (^-1/₂) = -1

Slope of p = 2

So we must choose the equation that has a slope of 2. If we rewrite the equations in point-slope form (y = mx + b), we see that the equation 2x – y = 3 could be written as y = 2x – 3. This means that the slope of the line 2x – y =3 would be 2, so it could be the equation of line p. The answer is 2x – y = 3.

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Example Question #1 : How To Find The Equation Of A Perpendicular Line

What is the equation for the line that is perpendicular to 4x - 3y = 6 through point (4, 6) ?

Possible Answers:

$y = \frac{-3}{4}x +2$

$y = \frac{4}{3}x + 6$

$y = \frac{4}{3}x - 5$

$y = \frac{1}{3}x - 3$

$y = \frac{-3}{4}x +9$

Correct answer:

$y = \frac{-3}{4}x +9$

Explanation:

Perpendicular slopes are opposite reciprocals.

The given slope is found by converting the equation to the slope-intercept form.

4x - 3y = 6

$y = \frac{4}{3}x - 2$

The slope of the given line is $m = \frac{4}{3}$ and the perpendicular slope is $m = \frac{-3}{4}$ .

We can use the given point and the new slope to find the perpendicular equation. Plug in the slope and the given coordinates to solve for the y-intercept.

$6 = \frac{-3}{4}(4) + b$

b = 9

Using this y-intercept in slope-intercept form, we get out final equation: $y = \frac{-3}{4}x + 9$ .

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Example Question #1 : How To Find The Equation Of A Perpendicular Line

Which line below is perpendicular to 5x+6y=18 ?

Possible Answers:

$y=-\frac{6}{5}x+8$

$y = \frac{6}{5}x + 3$

$y=\frac{5}{6}x+\frac{6}{5}$

$y=-\frac{5}{6}x+\frac{6}{5}$

$y=\frac{5}{6}x+2$

Correct answer:

$y = \frac{6}{5}x + 3$

Explanation:

The definition of a perpendicular line is one that has a negative, reciprocal slope to another.

For this particular problem, we must first manipulate our initial equation into a more easily recognizable and useful form: slope-intercept form or y=mx+b .

5x+6y=18

6y=-5x+18

$y=-\frac{5}{6}x+6$

According to our y=mx+b formula, our slope for the original line is $-\frac{5}{6}$ . We are looking for an answer that has a perpendicular slope, or an opposite reciprocal. The opposite reciprocal of $-\frac{5}{6}$ is $\frac{6}{5}$ . Flip the original and multiply it by .

Our answer will have a slope of $\frac{6}{5}$ . Search the answer choices for $\frac{6}{5}$ in the position of the y=mx+b equation.

$\dpi{100} y = \frac{6}{5}x + 3$ is our answer.

(As an aside, the negative reciprocal of 4 is $-\frac{1}{4}$ . Place the whole number over one and then flip/negate. This does not apply to the above problem, but should be understood to tackle certain permutations of this problem type where the original slope is an integer.)

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Example Question #1 : How To Find The Equation Of A Perpendicular Line

If a line has an equation of $2y=3x+3$ , what is the slope of a line that is perpendicular to the line?

Possible Answers:

$-\frac{3}{2}$

$-\frac{2}{3}$

$\frac{3}{2}$

$-2$

$3$

Correct answer:

$-\frac{2}{3}$

Explanation:

Putting the first equation in slope-intercept form yields $y=\frac{3}{2}x+\frac{3}{2}$ .

A perpendicular line has a slope that is the negative inverse. In this case, $-\frac{2}{3}$ .

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Example Question #1 : How To Find The Equation Of A Perpendicular Line

Given a line defined by the equation $y=\frac{9}{8}x+4$ , which of the following lines is perpendicular to ?

Possible Answers:

$y=-\frac{8}{9}x+4$

$y=\frac{9}{8}x+4$

$y=-\frac{9}{8}x+4$

$y=\frac{9}{8}x-4$

$y=4x+\frac{9}{8}$

Correct answer:

$y=-\frac{8}{9}x+4$

Explanation:

For a given line defined by the equation y=mx+b , any line perpendicular to must have a slope that is the negative reciprocal of 's slope , $-\frac{1}{m}$ .

In this instance, the slope of line is $m=\frac{9}{8}$ , so $-\frac{1}{m}=-\frac{1}{\frac{9}{8}}=-\frac{8}{9}$ . The only line provided with an equation that has this slope is $y=-\frac{8}{9}x+4$ .

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Example Question #2 : How To Find The Equation Of A Perpendicular Line

A given line is defined by the equation $y=\frac{5}{6}x-9$ . What is the slope of any line that is perpendicular to ?

Possible Answers:

Not enough information provided

$\frac{1}{9}$

$-\frac{6}{5}$

$-\frac{5}{6}$

Correct answer:

$-\frac{6}{5}$

Explanation:

For a given line defined by the equation y=mx+b , any line perpendicular to must have a slope that is the negative reciprocal of 's slope , $-\frac{1}{m}$ .

Since in this case $m=\frac{5}{6}$ ,

$-\frac{1}{m}=-\frac{1}{\frac{5}{6}}=\frac{6}{5}$ .

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Example Question #361 : Ssat Upper Level Quantitative (Math)

Which of the following equations represents a line that goes through the point (-2, 6) and is perpendicular to the line 3x + 6y = 12 ?

Possible Answers:

y = -2x + 2

y = 2x + 6

y = 2x + 2

y = -2x + 10

y = 2x + 10

Correct answer:

y = 2x + 10

Explanation:

In order to solve this problem, we need first to transform the equation from standard form to slope-intercept form:

y = mx + b

Transform the original equation to find its slope.

3x + 6y = 12

First, subtract from both sides of the equation.

3x-3x + 6y = 12-3x

Simplify and rearrange.

6y = -3x + 12

Next, divide both sides of the equation by 6.

$\frac{6y}{6} = -\frac{3x}{6} + \frac{12}{6}$

$y = -\frac{1}{2}x+ 2$

The slope of our first line is equal to $-\frac{1}{2}$ . Perpendicular lines have slopes that are opposite reciprocals of each other; therefore, if the slope of one is x, then the slope of the other is equal to the following:

$-\frac{1}{x}$

Let's calculate the opposite reciprocal of our slope:

$-\frac{1}{2}\rightarrow -\left (-\frac{2}{1} \right )=2$

The slope of our line is equal to 2. We now have the following partial equation:

y = 2x + b

We are missing the y-intercept, . Substitute the x- and y-values in the given point (-2, 6) to solve for the missing y-intercept.

6 = 2(-2) + b

6 = -4 + b

Add 4 to both sides of the equation.

6+4 = -4+4 + b

10 = b

Substitute this value into our partial equation to construct the equation of our line:

y = 2x + 10

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Example Question #1 : Distance Formula

A line segment has the endpoints (-1, 3) and (4,-2) . What is the length of this line segment?

Possible Answers:

$\sqrt{30}$

$5\sqrt2$

Correct answer:

$5\sqrt2$

Explanation:

Use the following formula to find the distance between two points:

$\text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Plug in the points that are given.

$\text{Distance}=\sqrt{(4-(-1))^2+(-2-3)^2}$

$\text{Distance}=\sqrt{(5)^2+(-5)^2}$

$\text{Distance}=\sqrt{25+25}=\sqrt{50}=\sqrt{5*5*2}=5\sqrt2$

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Example Question #372 : Ssat Upper Level Quantitative (Math)

A line segment has end points at (8,1) and (-2,-3) . What is the length of this line segment?

Possible Answers:

9.88

10.77

6.32

7.46

Correct answer:

10.77

Explanation:

The distance between two points is given by the following equation:

$\text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Now, using the two given points, plug them in to find the distance.

$\text{Distance}=\sqrt{(-2-8)^2+(-3-1)^2}=\sqrt{100+16}=\sqrt{116}=10.77$

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Example Question #93 : Coordinate Geometry

Find the length of a line that has end points at $(8, -1)\text{ and }(-2, 1)$ .

Possible Answers:

104

$2\sqrt{26}$

126

$\sqrt{26}$

Correct answer:

$2\sqrt{26}$

Explanation:

Use the distance formula to find the length of this line segment:

$\text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute in the values provided:

$\text{Distance}=\sqrt{(-2-8)^2+(1-(-1))^2}=\sqrt{100+4}=\sqrt{104}$

At this point, break down the square root:

$\sqrt{104}=\sqrt{2\cdot52}=\sqrt{2\cdot2\cdot26}=\sqrt{2\cdot2\cdot2\cdot13}$

You can remove two of the twos and put them on the outside of the square root symbol, and multiply the two and the thirteen that remain underneath the square root symbol:

$\sqrt{2\cdot2\cdot2\cdot13}=2\sqrt{2\cdot13}=2\sqrt{26}$

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All SSAT Upper Level Math Resources

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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 Equation Of A Perpendicular Line

Example Question #1 : How To Find The Equation Of A Perpendicular Line

Example Question #1 : How To Find The Equation Of A Perpendicular Line

Example Question #1 : How To Find The Equation Of A Perpendicular Line

Example Question #1 : How To Find The Equation Of A Perpendicular Line

Example Question #2 : How To Find The Equation Of A Perpendicular Line

Example Question #361 : Ssat Upper Level Quantitative (Math)

Example Question #1 : Distance Formula

Example Question #372 : Ssat Upper Level Quantitative (Math)

Example Question #93 : Coordinate Geometry

All SSAT Upper Level Math Resources

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