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SSAT Upper Level Math : How to find whether lines are parallel

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Example Questions

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 S

Line P

Line R

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 :Properties Of Parallel And Perpendicular Lines

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) .

这行re parallel?

Possible Answers:

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

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

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

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

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

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 #1 :Properties Of Parallel And Perpendicular Lines

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

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

Which statement is true of the two lines?

Possible Answers:

$\text{The lines are perpendicular.}$

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

$\text{The lines are parallel.}$

$\text{The lines are identical.}$

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

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 #1 :Properties Of Parallel And Perpendicular Lines

Parallel

Figure NOT drawn to scale

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

Possible Answers:

x = 168

x = 114

x = 132

x = 144

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 :Properties Of Parallel And Perpendicular Lines

Parallel

Figure NOT drawn to scale

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

Possible Answers:

y = 48-x

y = x+42

y = x-14

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 #421 :Geometry

Parallel

Figure NOT drawn to scale

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

Possible Answers:

y = 48-x

y = 138 - x

y = x-14

y = x+42

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 :Properties Of Parallel And Perpendicular Lines

Parallel

Figure NOT drawn to scale

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

Possible Answers:

x = 111

x = 106

x = 101

x= 96

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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SSAT Upper Level Math : How to find whether lines are parallel

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 :Properties Of Parallel And Perpendicular Lines

Example Question #1 :Properties Of Parallel And Perpendicular Lines

Example Question #1 :Properties Of Parallel And Perpendicular Lines

Example Question #5 :Properties Of Parallel And Perpendicular Lines

Example Question #421 :Geometry

Example Question #7 :Properties Of Parallel And Perpendicular Lines

All SSAT Upper Level Math Resources

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