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SSAT Upper Level Math : Properties of Parallel and Perpendicular Lines

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Example Questions

Example Question #11 : Properties Of Parallel And Perpendicular Lines

Which of the following lines is perpendicular to the line 3x-5y=11 ?

Possible Answers:

3x+5y=-11

3y-5x=-3

10y-6x+5=0

5y+3x=2

9y+15x-54=0

Correct answer:

9y+15x-54=0

Explanation:

All we care about for this problem is the slopes of the lines...the x- and y-intercepts are irrelevant.

Remember that the slopes of perpendicular lines are opposite reciprocals. By putting the given equation into y=mx+b form, we can see that its slope is $\frac{3}{5}$ . So we are looking for a line with a slope of $-\frac{5}{3}$ .

The equation 9y+15x-54=0 can be put into the form $y=-\frac{5}{3}x+6$ , and so we know that it is perpendicular to the given line.

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

Line A passes through the origin and (8, 6) .

Line B passes through the origin and $\left ( -36, 64\right )$ .

Line C passes through the origin and $\left ( \frac{1}{12},- \frac{1}{16} \right )$ .

Line D passes through the origin and (0.9, -1.2) .

Line E passes through the origin and $\left ( \sqrt{2},-\sqrt{3} \right )$ .

Which line is perpendicular to Line A?

Possible Answers:

Line C

None of the other lines is perpendicular to A.

Line D

Line B

Line E

Correct answer:

Line D

Explanation:

Find the slopes of all five lines 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 A:

$m = \frac{6 }{8 } = \frac{3}{4}$

The correct line must have as its slope the opposite of the reciprocal of this, which is $m = -\frac{4}{3}$ .

Line B:

$m = \frac{64 }{-36}= - \frac{16}{9}$

Line C:

$m = \frac{ -\frac{1}{16} }{ \frac{1}{12}} = -\frac{1}{16} \div \frac{1}{12} = -\frac{1}{16} \cdot \frac{12} {1}= -\frac{12}{16} = -\frac{3}{4}$

Line D:

$m = \frac{-1.2 }{0.9 } = -\frac{12}{9} = -\frac{4}{3}$

Line E:

$m = \frac{ - \sqrt{3}}{ \sqrt{2} } = -\frac{\sqrt{6}}{2}$

Of the last four lines, only Line D has the desired slope.

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

Line W passes through the origin and point $\left (\frac{1}{4}, \frac{1}{9} \right )$ .

Line X passes through the origin and point $\left (\frac{1}{4}, -\frac{1}{9} \right )$ .

Line Y passes through the origin and point $\left ( \frac{1}{9} , \frac{1}{4}\right )$ .

Line Z passes through the origin and point $\left ( \frac{1}{9} , -\frac{1}{4}\right )$ .

Which of these lines is perpendicular to the line of the equation 4x-9y = 100 ?

Possible Answers:

Line W

None of the other responses is correct.

Line X

Line Z

Line Y

Correct answer:

Line Z

Explanation:

First, find the slope of the line of the equation 4x-9y = 100 by rewriting it in slope-intercept form:

4x-9y = 100

-9y = -4x+100

$y =-\frac{1}{9} \left (-4x+100 \right )$

$y = \frac{4}{9} x-\frac{100 }{9}$

The slope of this line is $\frac{4}{9}$ , so we are looking for a line whose slope is the opposite of the reciprocal of this, or $-\frac{9}{4}$ .

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 W:

$m = \frac{y_{2} }{x_{2} }= \frac{ \frac{1}{9} }{ \frac{1}{4}} = \frac{1}{9} \div \frac{1}{4}= \frac{1}{9} \cdot \frac{4} {1} = \frac{4}{9}$

Line X:

$m = \frac{y_{2} }{x_{2} }= \frac{ - \frac{1}{9} }{ \frac{1}{4}} =- \frac{1}{9} \div \frac{1}{4}= -\frac{1}{9} \cdot \frac{4} {1} =- \frac{4}{9}$

Line Y:

$m = \frac{y_{2} }{x_{2} }= \frac{ \frac{1}{4} }{ \frac{1}{9}} = \frac{1}{4} \div \frac{1}{9}= \frac{1}{4} \cdot \frac{9} {1} = \frac{9}{4}$

Line Z:

$m = \frac{y_{2} }{x_{2} }= \frac{ -\frac{1}{4} }{ \frac{1}{9}} =-\frac{1}{4} \div \frac{1}{9}= -\frac{1}{4} \cdot \frac{9} {1} = -\frac{9}{4}$

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

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

Determine whether the two equations are parallel, perpendicular or neither, and choose the best reason.

$-2x+6y = 3 \textup{ and } -\frac{y}{3}=1$

Possible Answers:

Perpendicular, the slopes are the negative reciprocal to each other.

Parallel, the slopes are the same.

Neither, the slopes have no correlation.

Parallel, the slopes are the negative reciprocal to each other.

Perpendicular, the slopes are the same.

Correct answer:

Neither, the slopes have no correlation.

Explanation:

Convert both equations to slope intercept form: y=mx+b

-2x+6y = 3

6y =2x+3

$y=\frac{1}{3}x+\frac{1}{2}$

The slope of the first equation is $m_1=\frac{1}{3}$ .

Convert the second equation.

$-\frac{y}{3}=1$

y=-3

The slope of this equation is zero since there is no term! m_2=0

In order for the two functions to be parallel, they must have the same slopes.

In order for the two functions to be perpendicular, their slopes must be the negative reciprocal to each other.

Since there's no correlation with both slopes, the equations are neither parallel or perpendicular to each other.

The correct answer is:

Neither, the slopes have no correlation

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

Given: the following three lines on the coordinate plane:

Line 1: The line of the equation x = 7

Line 2: The line of the equation y = 7

Line 3: The line of the equation x+y=7

Which of the following is a true statement?

Possible Answers:

Line 1 and Line 2 are perpendicular; Line 3 is perpendicular to neither.

None of the other responses is correct.

Line 1 and Line 3 are perpendicular; Line 2 is perpendicular to neither.

Line 2 and Line 3 are perpendicular; Line 1 is perpendicular to neither.

No two of Line 1, Line 2, or Line 3 form a pair of perpendicular lines.

Correct answer:

Line 1 and Line 2 are perpendicular; Line 3 is perpendicular to neither.

Explanation:

Line 1, the line of the equation x = 7 , is a vertical line on the coordinate plane; Line 2, the line of the equation y = 7 , is a horizontal line. Lines 1 and 2 are perpendicular to each other.

The slope of Line 3, the line of the equation x+y=7 , can be calculated by putting the equation in slope-intercept form:

x+y=7

x+y-x=7 - x

y = -1x+7

The slope is , which makes it perpendicular to a line of slope $-\frac{1}{-1} = 1$ . Line 1, being vertical, has undefined slope, and Line 2, being horizontal, has slope 0.

Correct response: Line 1 and Line 2 are perpendicular; Line 3 is perpendicular to neither.

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Example Question #13 : Properties Of Parallel And Perpendicular Lines

The line of the equation 6x - 8y = 17 is perpendicular to which of the following lines on the coordinate plane?

Possible Answers:

$y = \frac{4}{3}x$

$y =- \frac{4}{3}x$

$y = \frac{3}{4}x$

$y =- \frac{3}{4}x$

None of the other responses is correct.

Correct answer:

$y =- \frac{4}{3}x$

Explanation:

First, find the slope of the line 6x - 8y = 17 by rewriting the equation in slope-intercept form and noting the coefficient of :

6x - 8y = 17

6x - 8y-6 x = 17-6 x

- 8y = -6x+17

$\frac{- 8y}{-8} =\frac{ -6x+17}{-8}$

$y = \frac{3}{4}x-\frac{17}{8}$

The line has slope $\frac{3}{4}$ .

A line perpendicular to this would have slope $-1 \div \frac{3}{4} = -\frac{4}{3}$ . Of the four equations among the choices, all of which are in slope-intercept form, only $y =- \frac{4}{3}x$ has this slope.

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

One side of a rectangle on the coordinate plane has as its endpoints the points (2, 4) and (9, -1) .

What would be the slope of a side adjacent to this side?

Possible Answers:

$\frac{7}{5}$

$\frac{5}{7}$

None of the other responses gives the correct answer.

$\frac{3}{7}$

$\frac{7}{3}$

Correct answer:

$\frac{7}{5}$

Explanation:

First, we find the slope of the segment connecting (2, 4) or (9, -1) . Using the formula

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

and setting

$x_{1} = 2, y_{1}=4, x_{2}= 9, y_{2} = -1$

we get

$m = \frac{-1-4}{9-2}= \frac{-5}{7} = -\frac{5}{7}$

Adjacent sides of a rectangle are perpendicuar, so their slopes will be the opposites of each other's reciprocals. Therefore, the slope of an adjacent side will be the opposite of the reciprocal of $-\frac{5}{7}$ , which is $\frac{7}{5}$ .

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Example Question #11 : How To Find Whether Lines Are Perpendicular

Lines 1 and 2, which are perpendicular, have their -intercepts at the point (0, 8) . The -intercept of Line 1 is at the point (5,0) . Give the -intercept of Line 2.

Possible Answers:

$\left ( \frac{5 }{64}, 0\right )$

$\left ( 12 \frac{4}{5}, 0 \right )$

$\left ( -12 \frac{4}{5}, 0 \right )$

$\left (-\frac{5 }{64}, 0\right )$

(-5,0)

Correct answer:

$\left ( -12 \frac{4}{5}, 0 \right )$

Explanation:

The slope of a line with -intercept (a, 0) and -intercept (0, b) is $m = -\frac {b}{a}$ . For Line 1, b= 8, a= 5 , so Line 1 has slope $-\frac{8}{5}$ . The slope of Line 2, which is perpendicular to Line 1, will be the opposite of the reciprocal of this, which is $\frac{5}{8}$ . Setting equal to this and b=8 , we get

$\frac{5}{8}= -\frac {8}{a}$ , or $\frac{5}{8}= \frac {-8}{a}$

Cross-multiplying:

5a = -64

$\frac{5a }{5}= \frac{-64}{5}$

$a = -12 \frac{4}{5}$

The -intercept of Line 2 is $\left ( -12 \frac{4}{5}, 0 \right )$ .

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Example Question #12 : How To Find Whether Lines Are Perpendicular

Which of the following choices gives the equations of a pair of perpendicular lines with the same -intercept?

Possible Answers:

y = 2x+ 6 and $y = \frac{1}{2}x + 6$

y = 4x-8 and $y = - \frac{1}{4}x + \frac{1}{8}$

$y = \frac{2}{5}x + 9$ and $y = - \frac{5}{2}x + 9$

y = 3x+ 7 and y = -3x + 7

$y = \frac{1}{3}x + \frac{2}{3}$ and $y = - \frac{1}{3}x - \frac{2}{3}$

Correct answer:

$y = \frac{2}{5}x + 9$ and $y = - \frac{5}{2}x + 9$

Explanation:

All of the equations are given in slope-intercept form y = mx+b , so we can answer this question by examining the coefficients of , which are the slopes, and the constants, which are the -intercepts. In each case, since the lines are perpendicular, each -coefficient must be the other's opposite reciprocal, and since the lines have the same -intercept, the constants must be equal.

Of the five pairs, only

y = 4x-8 and $y = - \frac{1}{4}x + \frac{1}{8}$

and

$y = \frac{2}{5}x + 9$ and $y = - \frac{5}{2}x + 9$

have equations whose -coefficients are the other's opposite reciprocal. Of these, only the latter pair of equations have equal constant terms.

$y = \frac{2}{5}x + 9$ and $y = - \frac{5}{2}x + 9$

is the correct choice.

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Example Question #13 : How To Find Whether Lines Are Perpendicular

Given: the following three lines on the coordinate plane:

Line 1: The line of the equation x+2y = 12

Line 2: The line of the equation 2x+y= 12

Line 3: The line of the equation y=2 x

Which of the following is a true statement?

Possible Answers:

None of the other responses is correct.

No two of Line 1, Line 2, or Line 3 form a pair of perpendicular lines.

Line 1 and Line 3 are perpendicular; Line 2 is perpendicular to neither.

Line 2 and Line 3 are perpendicular; Line 1 is perpendicular to neither.

Line 1 and Line 2 are perpendicular; Line 3 is perpendicular to neither.

Correct answer:

Line 1 and Line 3 are perpendicular; Line 2 is perpendicular to neither.

Explanation:

The slope of each line can be calculated by putting the equation in slope-intercept form y = mx+b and noting the coefficient of :

Line 1:

x+2y = 12

x+2y - x = 12 - x

2y = -1x + 12

$\frac{2y }{2}= \frac{-1x + 12}{2}$

$y= -\frac{1}{2}x+6$

Slope of Line 1: $-\frac{1}{2}$

Line 2:

2x+y= 12

2x+y - 2x= 12 - 2x

y = -2x+12

Slope of Line 2:

Line 3: The equation is already in slope-intercept form; its slope is 2.

Two lines are perpendicular if and only their slopes have product . The slopes of Lines 1 and 3 have product $-\frac{1}{2} \cdot 2 = -1$ ; they are perpendicular. The slopes of Lines 1 and 2 have product $-\frac{1}{2} \cdot (- 2 )= 1$ ; they are not perpendicular. The slopes of Lines 2 and 3 have product $-2 \cdot 2 = -4$ ; they are not perpendicular.

Correct response: Line 1 and Line 3 are perpendicular; Line 2 is perpendicular to neither.

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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 #11 : Properties Of Parallel And Perpendicular Lines

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

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

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

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

Example Question #13 : Properties Of Parallel And Perpendicular Lines

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

Example Question #11 : How To Find Whether Lines Are Perpendicular

Example Question #12 : How To Find Whether Lines Are Perpendicular

Example Question #13 : How To Find Whether Lines Are Perpendicular

All SSAT Upper Level Math Resources

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