How to find whether lines are parallel - SSAT Upper Level Quantitative

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Question

Line A has equation .

Line B has equation .

Which statement is true of the two lines?

Answer

Write each statement in slope-intercept form:

Line A:

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

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

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

Line B:

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

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

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

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

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Question

You are given three lines as follows:

Line A includes points and .

Line B includes point and has -intercept .

Line C includes the origin and point .

Which lines are parallel?

Answer

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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Question

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 ?

Answer

First, find the slope of the line of the equation by rewriting it in slope-intercept form:

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

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

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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Question

Figure NOT drawn to scale

In the above figure, . Evaluate .

Answer

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,

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Question

Figure NOT drawn to scale

In the above figure, . Express in terms of .

Answer

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

Solving for by subtracting 28 from both sides:

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Question

Figure NOT drawn to scale

In the above figure, . Express in terms of .

Answer

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,

Solve for by moving the other terms to the other side and simplifying:

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Question

Figure NOT drawn to scale

In the above figure, . Evaluate .

Answer

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

Solving for :

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

The angles of measures and form a pair of alternating interior angles of parallel lines, so, as a consequence of the Parallel Postulate, they are congruent, and

Substituting for :

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Question

Line A has equation .

Line B has equation .

Which statement is true of the two lines?

Answer

Write each statement in slope-intercept form:

Line A:

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

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

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

Line B:

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

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

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

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

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Question

You are given three lines as follows:

Line A includes points and .

Line B includes point and has -intercept .

Line C includes the origin and point .

Which lines are parallel?

Answer

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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Question

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 ?

Answer

First, find the slope of the line of the equation by rewriting it in slope-intercept form:

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

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

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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Question

Figure NOT drawn to scale

In the above figure, . Evaluate .

Answer

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,

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Question

Figure NOT drawn to scale

In the above figure, . Express in terms of .

Answer

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

Solving for by subtracting 28 from both sides:

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Question

Figure NOT drawn to scale

In the above figure, . Express in terms of .

Answer

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,

Solve for by moving the other terms to the other side and simplifying:

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Question

Figure NOT drawn to scale

In the above figure, . Evaluate .

Answer

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

Solving for :

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

The angles of measures and form a pair of alternating interior angles of parallel lines, so, as a consequence of the Parallel Postulate, they are congruent, and

Substituting for :

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Question

Line A has equation .

Line B has equation .

Which statement is true of the two lines?

Answer

Write each statement in slope-intercept form:

Line A:

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

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

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

Line B:

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

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

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

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

Compare your answer with the correct one above

Question

You are given three lines as follows:

Line A includes points and .

Line B includes point and has -intercept .

Line C includes the origin and point .

Which lines are parallel?

Answer

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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Question

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 ?

Answer

First, find the slope of the line of the equation by rewriting it in slope-intercept form:

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

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

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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Question

Figure NOT drawn to scale

In the above figure, . Evaluate .

Answer

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,

Compare your answer with the correct one above

Question

Figure NOT drawn to scale

In the above figure, . Express in terms of .

Answer

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

Solving for by subtracting 28 from both sides:

Compare your answer with the correct one above

Question

Figure NOT drawn to scale

In the above figure, . Express in terms of .

Answer

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,

Solve for by moving the other terms to the other side and simplifying:

Compare your answer with the correct one above

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