How to find whether lines are perpendicular

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SSAT Upper Level Quantitative › How to find whether lines are perpendicular

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

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

CORRECT

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

and

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

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

Explanation

All of the equations are given in slope-intercept form , 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

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.

One side of a rectangle on the coordinate plane has as its endpoints the points and .

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

$\frac{7}{5}$

CORRECT

$\frac{5}{7}$

$\frac{3}{7}$

$\frac{7}{3}$

None of the other responses gives the correct answer.

Explanation

First, we find the slope of the segment connecting or . 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 $\frac{7}{5}$ .

Line A passes through the origin and .

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 .

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

Which line is perpendicular to Line A?

Line D

CORRECT

Line B

Line C

Line E

None of the other lines is perpendicular to A.

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.

The line of the equation is perpendicular to which of the following lines on the coordinate plane?

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

CORRECT

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

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

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

None of the other responses is correct.

Explanation

First, find the slope of the line by rewriting the equation in slope-intercept form and noting the coefficient of :

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

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

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.

Given: the following three lines on the coordinate plane:

Line 1: The line of the equation

Line 2: The line of the equation

Line 3: The line of the equation

Which of the following is a true statement?

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

CORRECT

Line 1 and Line 2 are perpendicular; Line 3 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.

None of the other responses is correct.

Explanation

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

Line 1:

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

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

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

Line 2:

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.

Given: the following three lines on the coordinate plane:

Line 1: The line of the equation

Line 2: The line of the equation

Line 3: The line of the equation

Which of the following is a true statement?

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

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.

None of the other responses is correct.

Explanation

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

The slope of Line 3, the line of the equation , can be calculated by putting the equation in slope-intercept form:

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.

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 ?

Line Z

CORRECT

Line Y

Line X

Line W

None of the other responses is correct.

Explanation

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

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

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

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.

Three lines are drawn on the coordinate plane.

The green line has slope $\frac{2}{3}$ , and -intercept $\frac{4}{5}$ .

The blue line has slope $\frac{3}{2}$ , and -intercept $-\frac{5}{4}$ .

The red line has slope $-\frac{2}{3}$ , and -intercept $-\frac{4}{5}$ .

Which two lines are perpendicular to each other?

The blue line and the red line are perpendicular.

CORRECT

The blue line and the green line are perpendicular.

The green line and the red line are perpendicular.

No two of these lines are perpendicular.

It cannot be determined from the information given.

Explanation

To demonstrate two perpendicular lines, multiply their slopes; if their product is , then the lines are perpendicular (the -intercepts are irrelevant).

The products of these lines are given here.

Blue and green lines: $\frac{2}{3} * \frac{3}{2} = 1$

Red and green lines: $\frac{2}{3} * \left - \frac{2}{3} \right = -\frac{4}{9}$

Blue and red lines: $\frac{2}{3}* \left - \frac{3}{2} \right = -1$

It is the blue and red lines that are perpendicular.

We can also see that their slopes are negative reciprocals, indicating perpendicular lines.

Two perpendicular lines intersect at the origin; one line also passes through point . What is the slope of the other line?

$-\frac{5}{ 7}$

CORRECT

$-\frac{ 7}{5}$

$\frac{5}{ 7}$

$\frac{ 7}{5}$

Insufficient information is given to solve the problem.

Explanation

The slopes of two perpendicular lines are the opposites of each other's reciprocals.

To find the slope of the first line, substitute $x_{1} = 5, y_{1} =7, x_{2} = 0, y_{2} =0$ in the slope formula:

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

$m =\frac{ 7-0}{5-0}=\frac{ 7}{5}$

The slope of the first line is $\frac{ 7}{5}$ , so the slope of the second line is the opposite reciprocal of this, which is $-\frac{5}{ 7}$ .

Which of the following lines is perpendicular to the line ?

CORRECT

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 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 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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