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

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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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