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SSAT Upper Level Math : Properties of Parallel and Perpendicular Lines

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Example Question #1 :Properties Of Parallel And Perpendicular Lines

Line P passes through the origin and point $\left ( \frac{1}{9} , \frac{1}{7}\right )$ .

Line Q passes through the origin and point $\left ( \frac{1}{9} , -\frac{1}{7}\right )$ .

Line R passes through the origin and point $\left (\frac{1}{7}, \frac{1}{9} \right )$ .

Line S passes through the origin and point $\left (\frac{1}{7}, -\frac{1}{9} \right )$ .

Which of these lines is parallel to the line of the equation 7x+9y = 111 ?

Possible Answers:

Line Q

Line R

Line P

Line S

None of the other responses is correct.

Correct answer:

Line S

Explanation:

First, find the slope of the line of the equation 7x+9y = 111 by rewriting it in slope-intercept form:

7x+9y = 111

9y =-7x+ 111

$y =\frac{1}{9} \left (-7x+ 111 \right )$

$y = -\frac{7}{9}x+ \frac{111 }{9}$

The slope of this line is $-\frac{7}{9}$ , so we are looking for a line which also has this slope.

Find the slopes of all four lines by using the slope formula $m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ . Since each line passes through the origin, this formula can be simplified to

$m= \frac{y_{2}-0}{x_{2}-0} = \frac{y_{2} }{x_{2} }$

using the other point.

Line P:

$m = \frac{y_{2} }{x_{2} }= \frac{ \frac{1}{7} }{ \frac{1}{9}} = \frac{1}{7} \div \frac{1}{9}= \frac{1}{7} \cdot \frac{9} {1} = \frac{9}{7}$

Line Q:

$m = \frac{y_{2} }{x_{2} }= \frac{- \frac{1}{7} }{ \frac{1}{9}} =- \frac{1}{7} \div \frac{1}{9}= -\frac{1}{7} \cdot \frac{9} {1} =- \frac{9}{7}$

Line R:

$m = \frac{y_{2} }{x_{2} }= \frac{ \frac{1}{9} }{ \frac{1}{7}} = \frac{1}{9} \div \frac{1}{7}= \frac{1}{9} \cdot \frac{7} {1} = \frac{7}{9}$

Line S:

$m = \frac{y_{2} }{x_{2} }= \frac{- \frac{1}{9} }{ \frac{1}{7}} = -\frac{1}{9} \div \frac{1}{7}=- \frac{1}{9} \cdot \frac{7} {1} =-\frac{7}{9}$

Line S has the desired slope and is the correct choice.

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Example Question #2 :How To Find Whether Lines Are Parallel

You are given three lines as follows:

Line A includes points (3, 8) and (1, 12) .

Line B includes point (6, 7) and has-intercept (0, 17) .

Line C includes the origin and point (4, -8) .

Which lines are parallel?

Possible Answers:

$\text{Only Lines B and C are parallel.}$

$\text{Only Lines A and C are parallel.}$

$\text{No two of Lines A, B, and C are parallel.}$

$\text{All three of Lines A, B, and C are parallel.}$

$\text{Only Lines A and B are parallel.}$

Correct answer:

$\text{Only Lines A and C are parallel.}$

Explanation:

Find the slope of all three lines using the slope formula $m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ :

Line A:

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

$m= \frac{12-8}{1-3} = \frac{4}{-2} = -2$

Line B:

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

$m= \frac{7-17}{6-0} = \frac{-10}{6} = -\frac{5}{3}$

Line C:

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

$m= \frac{-8-0}{4-0}= \frac{-8}{4} = -2$

Lines A and C have the same slope; Line B has a different slope. Only Lines A and C are parallel.

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Example Question #421 :Geometry

Line A has equation 5x-7y = 20 .

Line B has equation 7x-5y = 32 .

Which statement is true of the two lines?

Possible Answers:

$\text{The lines are perpendicular.}$

$\text{The lines are identical.}$

$\text{None of the other four statements are true.}$

$\text{The lines are parallel.}$

$\text{The lines are different but neither parallel nor perpendicular.}$

Correct answer:

$\text{The lines are different but neither parallel nor perpendicular.}$

Explanation:

Write each statement in slope-intercept form:

Line A:

5x-7y = 20

-7y = -5x+ 20

$y =-\frac{1}{7}\left ( -5x+ 20 \right )$

$y = \frac{5}{7} x-\frac{ 20}{7}$

The slope is $\frac{5}{7}$ .

Line B:

7x-5y = 32

-5y = - 7x+ 32

$y = -\frac{1}{5 }\left ( - 7x+ 32 \right )$

$y = \frac{7}{5}x-\frac{32}{5}$

The slope is $\frac{7}{5}$ .

The lines have differing slopes, so they are neither identical nor parallel. The product of the slopes is $\frac{5}{7} \times \frac{7}{5} = 1 \ne -1$ , so they are not perpendicular. The correct response is that they are distinct lines that are neither parallel nor perpendicular.

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Example Question #4 :How To Find Whether Lines Are Parallel

Parallel

Figure NOT drawn to scale

In the above figure, m ||n . Evaluate.

Possible Answers:

x = 168

x = 114

x = 144

x = 132

Correct answer:

x = 132

Explanation:

Angles of degree measuresand 3y - 12 form a linear pair, making the angles supplementary - that is, their degree measures total 180. Therefore,

y + (3y - 12) = 180

Solving for:

4 y - 12 = 180

4 y - 12 + 12 = 180 + 12

4y = 192

$4y \div 4 = 192 \div 4$

y = 48

The angles of measuresand 3y - 12 form a pair of alternating interior angles of parallel lines, so, as a consequence of the Parallel Postulate, they are congruent, and

x = 3y - 12

Substituting for:

x = 3(48) - 12 = 144 - 12 = 132

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Example Question #5 :How To Find Whether Lines Are Parallel

Parallel

Figure NOT drawn to scale

In the above figure, m ||n . Expressin terms of.

Possible Answers:

y = x-14

y = x+42

y = 48-x

y = 138 - x

Correct answer:

y = 138 - x

Explanation:

The two marked angles are same-side interior angles of two parallel lines formed by a transversal; by the Parallel Postulate, the angles are supplementary - the sum of their measures is 180 degrees. Therefore,

(y+28 )+ ( x+ 14) = 180

x+y + 42 = 180

Solve forby moving the other terms to the other side and simplifying:

x+y + 42- 42 - x = 180 - 42 - x

y = 138 - x

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Example Question #6 :How To Find Whether Lines Are Parallel

Parallel

Figure NOT drawn to scale

In the above figure, m ||n . Expressin terms of.

Possible Answers:

y = 138 - x

y = x+42

y = 48-x

y = x-14

Correct answer:

y = x-14

Explanation:

The two marked angles are corresponding angles of two parallel lines formed by a transversal, so the angles are congruent. Therefore,

y+28 = x+ 14

Solving forby subtracting 28 from both sides:

y+28 - 28 = x+ 14 - 28

y = x-14

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Example Question #7 :How To Find Whether Lines Are Parallel

Parallel

Figure NOT drawn to scale

In the above figure, m ||n . Evaluate.

Possible Answers:

x= 96

x = 101

x = 111

x = 106

Correct answer:

x = 101

Explanation:

The two marked angles are same-side exterior angles of two parallel lines formed by a transversal,; by the Parallel Postulate, the angles are supplementary - the sum of their measures is 180 degrees. Therefore,

(x+12) + (x- 34) = 180

2x -22 = 180

2x = 202

x = 101

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Example Question #1 :Properties Of Parallel And Perpendicular Lines

Three lines are drawn on the coordinate plane.

The green line has slope $\frac{2}{3}$ , and-intercept $\frac{4}{5}$ .

The blue line has slope $\frac{3}{2}$ , and-intercept $-\frac{5}{4}$ .

The red line has slope $-\frac{2}{3}$ , and-intercept $-\frac{4}{5}$ .

Which two lines are perpendicular to each other?

Possible Answers:

The blue line and the green line are perpendicular.

It cannot be determined from the information given.

No two of these lines are perpendicular.

The green line and the red line are perpendicular.

The blue line and the red line are perpendicular.

Correct answer:

The blue line and the red line are perpendicular.

Explanation:

To demonstrate two perpendicular lines, multiply their slopes; if their product is, then the lines are perpendicular (the-intercepts are irrelevant).

The products of these lines are given here.

Blue and green lines: $\frac{2}{3} * \frac{3}{2} = 1$

Red and green lines: $\frac{2}{3} * \left - \frac{2}{3} \right = -\frac{4}{9}$

Blue and red lines: $\frac{2}{3}* \left - \frac{3}{2} \right = -1$

It is the blue and red lines that are perpendicular.

We can also see that their slopes are negative reciprocals, indicating perpendicular lines.

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Example Question #1 :How To Find Whether Lines Are Perpendicular

Two perpendicular lines intersect at point (3,8) . One line also includes point (4,-1) . What is the slope of theotherline?

Possible Answers:

Insufficient information is given to answer the question.

$\frac{1}{9}$

$-\frac{1}{9}$

Correct answer:

$\frac{1}{9}$

Explanation:

The slopes of two perpendicular lines are the opposites of each other's reciprocals.

To find the slope of the first line substitute $x_{1} = 3, y_{1} =8, x_{2} = 4, y_{2} =-1$ in the slope formula:

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

$m =\frac{ 8 - (-1)}{ 3-4} = \frac{ 9}{ -1} = -9$

The slope of the first line is, so the slope of the second line is the opposite reciprocal of this, which is $\frac{1}{9}$ .

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Example Question #3 :Properties Of Parallel And Perpendicular Lines

Two perpendicular lines intersect at the origin; one line also passes through point (5,7) . What is the slope of the other line?

Possible Answers:

$-\frac{5}{ 7}$

$\frac{5}{ 7}$

$\frac{ 7}{5}$

Insufficient information is given to solve the problem.

$-\frac{ 7}{5}$

Correct answer:

$-\frac{5}{ 7}$

Explanation:

The slopes of two perpendicular lines are the opposites of each other's reciprocals.

To find the slope of the first line, substitute $x_{1} = 5, y_{1} =7, x_{2} = 0, y_{2} =0$ in the slope formula:

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

$m =\frac{ 7-0}{5-0}=\frac{ 7}{5}$

The slope of the first line is $\frac{ 7}{5}$ , so the slope of the second line is the opposite reciprocal of this, which is $-\frac{5}{ 7}$ .

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SSAT Upper Level Math : Properties of Parallel and Perpendicular Lines

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

All SSAT Upper Level Math Resources

Example Questions

Example Question #1 :Properties Of Parallel And Perpendicular Lines

Example Question #2 :How To Find Whether Lines Are Parallel

Example Question #421 :Geometry

Example Question #4 :How To Find Whether Lines Are Parallel

Example Question #5 :How To Find Whether Lines Are Parallel

Example Question #6 :How To Find Whether Lines Are Parallel

Example Question #7 :How To Find Whether Lines Are Parallel

Example Question #1 :Properties Of Parallel And Perpendicular Lines

Example Question #1 :How To Find Whether Lines Are Perpendicular

Example Question #3 :Properties Of Parallel And Perpendicular Lines

All SSAT Upper Level Math Resources

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