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

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Line A has equation .

Line B has equation .

Which statement is true of the two lines?

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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 \ne -1$ , so they are not perpendicular. The correct response is that they are distinct lines that are neither parallel nor perpendicular.

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?

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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.

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 ?

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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.

Figure NOT drawn to scale

In the above figure, . Evaluate .

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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,

Figure NOT drawn to scale

In the above figure, . Express in terms of .

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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:

Figure NOT drawn to scale

In the above figure, . Express in terms of .

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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:

Figure NOT drawn to scale

In the above figure, . Evaluate .

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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 :

Line A has equation .

Line B has equation .

Which statement is true of the two lines?

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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 \ne -1$ , so they are not perpendicular. The correct response is that they are distinct lines that are neither parallel nor perpendicular.

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?

Tap to see back →

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.

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 ?

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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.

Figure NOT drawn to scale

In the above figure, . Evaluate .

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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,

Figure NOT drawn to scale

In the above figure, . Express in terms of .

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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:

Figure NOT drawn to scale

In the above figure, . Express in terms of .

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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:

Figure NOT drawn to scale

In the above figure, . Evaluate .

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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 :

Line A has equation .

Line B has equation .

Which statement is true of the two lines?

Tap to see back →

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 \ne -1$ , so they are not perpendicular. The correct response is that they are distinct lines that are neither parallel nor perpendicular.

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?

Tap to see back →

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.

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 ?

Tap to see back →

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.

Figure NOT drawn to scale

In the above figure, . Evaluate .

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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,

Figure NOT drawn to scale

In the above figure, . Express in terms of .

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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:

Figure NOT drawn to scale

In the above figure, . Express in terms of .

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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: