How to find out if lines are perpendicular - SAT Math

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Question

Which of the following lines is perpendicular to the line ?

Answer

Perpendicular lines will have slopes that are negative reciprocals of one another. Our first step will be to find the slope of the given line by putting the equation into slope-intercept form.

The slope of this line is . The negative reciprocal will be $+\frac{1}{3}$ , which will be the slope of the perpendicular line.

Now we need to find the answer choice with this slope by converting to slope-intercept form.

$y=\frac{4}{12}x-\frac{6}{12}$

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

This equation has a slope of $+\frac{1}{3}$ , and must be our answer.

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Question

Which set of lines is perpendicular?

Answer

Two lines are perpendicular to each other if their slopes are negative reciprocals. For example, if one line has a slope of 2, the line perpendicular to it has a slope of –1/2. One easy way to eliminate answer choices is to check if the slopes have the same sign, i.e. both positive or both negative. If so, they cannot be perpendicular. Several of the lines in the answer choices are of the form y = mx + b, where m is the slope and b is the y-intercept. We are only worried about the slope for the purposes of this question.

y = 3_x_ + 5 and y = 5_x_ + 3 both have positive slopes (m = 3 and m = 5, respectively), so they aren't perpendicular.

y = 3_x_/5 – 3 and y = 5_x_/3 + 3 both have positive slopes, so again they aren't perpendicular.

y = x – 1/2 and y = –x + 1/2 have slopes of m = 1 and m = –1, respectively. One is positive and one is negative, so that is a good sign. Let's take the negative reciprocal of 1. 1 /–1 = –1. So these two slopes are in fact negative reciprocals, and these two lines are perpendicular to each other. Even though we have found the correct answer, let's go through the other two choices to be sure.

The line between the points (1,3) and (3,5), and y = 4_x_ + 7: We need to find the slope of the first line. slope = rise / run = (_y_2 – _y_1) / (_x_2 – x_1) = (5 – 3) / (3 – 1) = 1. The slope of y = 4_x + 7 is also positive (m = 4), so the lines are not perpendicular.

The line between the points (7,4) and (4,7), and the line between the points (3,9) and (4,8): the first slope = (7 – 4) / (4 – 7) = –1 and the second slope = (8 – 9) / (4 – 3) = –1. They have the same slope, making them parallel, not perpendicular.

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Question

If two lines have slopes of $-5$ and $\frac{1}{5}$ , which statement about the lines is true?

Answer

Perpendicular lines have slopes that are the negative reciprocals of each other.

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Question

Which of the following lines is perpendicular to $y=3x-4$

Answer

The line which is perpendicular has a slope which is the negative inverse of the slope of the original line.

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Question

Which of the following equations represents a line that is perpendicular to the line with points and ?

Answer

If lines are perpendicular, then their slopes will be negative reciprocals.

First, we need to find the slope of the given line.

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{11-(-13)}{7-5}$

$m=\frac{24}{2}$

Because we know that our given line's slope is , the slope of the line perpendicular to it must be $-\frac{1}{12}$ .

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Question

Which of the following lines is perpindicular to

Answer

When determining if a two lines are perpindicular, we are only concerned about their slopes. Consider the basic equation of a line, , where m is the slope of the line. Two lines are perpindicular to each other if one slope is the negative and reciprocal of the other.

The first step of this problem is to get it into the form, , which is . Now we know that the slope, m, is . The reciprocal of that is $\frac{1}{3}$ , and the negative of that is $\frac{-1}{3}$ . Therefore, any line that has a slope of $\frac{-1}{3}$ will be perpindicular to the original line.

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Question

Which of the following lines is NOT perpendicular to ?

Answer

Perpendicular lines have slopes that are negative reciprocals. The given equation has a slope of ; therefore, any lime that is perpendicular to it will have a slope of $- \frac {1}{3}$ .

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Question

Line is given by $-\frac{6}{5}x+y=4$ .

Which of the following is perpendicular to ?

Answer

Putting the equation of the line into slope-intercept form, we get

$y=\frac{6}{5}x+4$

The slope of line , therefore, is $\frac{6}{5}$ .

In order for a line to be perpendicular to the given line, it must have a slope that is the negative reciprocal of line g's slope.

The slope of any given line perpendicular to line g must be $-\frac{5}{6}$ when written in slope-intercept form. In other words, the equation of the perpendicular line must be $y=-\frac{5}{6}x +k$ where k is any constant.

Written in standard form, the equation of this perpendicular line must be $\frac{5}{6}x+y=k.$

Therefore, the most appropriate answer is $\frac{5}{6}x+y=6.$

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Question

Given: Lines A, B, and C on the coordinate plane, as follows:

The equation of Line A is .

The equation of Line B is .

The equation of Line C is .

Which of the following is a true statement?

Answer

Two lines are perpendicular if and only if the product of their slopes is . Therefore, we need to find the slopes of all three lines.

Rewrite the equation for each line in its slope-intercept form , where is the slope of the line.

Line A:

Isolate by working the same operations on both sides:

$\frac{1}{7} \cdot 7y = \frac{1}{7} \cdot ( - 2x+ 13)$

$y = \frac{1}{7} \cdot ( - 2x) +\frac{1}{7} \cdot 13$

$y = - \frac{2}{7} x +\frac{13}{7}$

The slope of Line A is the coefficient of , which is $- \frac{2}{7}$ .

Take the same steps with the equations of the other two lines:

Line B:

$- \frac{1}{2}(- 2y) =- \frac{1}{2}( -7x + 13)$

$y = - \frac{1}{2}( -7x) +\left ( - \frac{1}{2} \right ) ( 13)$

$y = \frac{7}{2} x - \frac{13}{2}$

The slope of Line B is $\frac{7}{2}$

Line C:

$-\frac{1}{7}(-7y) = -\frac{1}{7} (- 2x + 13)$

$y = -\frac{1}{7} (- 2x ) + \left ( -\frac{1}{7} \right )(13)$

$y = \frac{2}{7} x -\frac{13}{7}$

The slope of Line C is $\frac{2}{7}$ .

The product of the slopes of Lines A and B is $- \frac{2}{7} \times \frac{7}{2} = -1$ , so these two lines are perpendicular.

The product of the slopes of Lines A and C is $- \frac{2}{7} \times \frac{2}{7} = -\frac{4}{49}$ , so these two lines are not perpendicular.

The product of the slopes of Lines B and C is $\frac{7}{2} \times \frac{2}{7}= 1$ , so these two lines are not perpendicular.

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Question

Given: Lines A, B, and C on the coordinate plane, as follows:

The equation of Line A is .

The equation of Line B is .

The equation of Line is .

Which of the following is a true statement?

Answer

Two lines are perpendicular if and only if the product of their slopes is . Therefore, we need to find the slopes of all three lines.Rewrite the equation for each line in its slope-intercept form , where is the slope of the line.

Line A:

is already in this form. The slope of Line A is the coefficient of , which is 3.

Line B:

Isolate by working the same operations on both sides:

$\frac{1}{3} \cdot 3y =\frac{1}{3} \cdot ( 9x + 17)$

$y =\frac{1}{3} \cdot 9x +\frac{1}{3} \cdot 17$

$y =3 x +\frac{17}{3}$

The slope of Line B is the coefficient of , which is 3.

Line C:

$- \frac{1}{2}(- 2y) = - \frac{1}{2} (- 6x + 17)$

$y = - \frac{1}{2} (- 6x) + \left ( - \frac{1}{2} \right )(17)$

$y = 3x - \frac{17}{2}$

The slope of Line C is the coefficient of , which is 3.

All three lines have slope 3, so the product of the slopes of any two of the lines is $3 \times 3 = 9 e -1$ . Therefore, no two of the lines are perpendicular.

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Question

Given: Lines A, B, and C on the coordinate plane, as follows:

Line A has intercepts and .

Line B has intercepts and .

Line C has intercepts and .

Which of the following statements is true?

Answer

The slope of each line, given the coordinates of two points through which they pass, can be calculated by substituting the point coordinates into the slope formula

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

Line A:

Setting $y_{1} = x_{2}= 0, x_{1} = 6 , y_{2} = 4$ and substituting:

$m = \frac{4-0}{0- 6} = \frac{4}{-6} = -\frac{2}{3}$

Line B:

Setting $y_{1} = x_{2}= 0, x_{1} = 9 , y_{2} = -6$ and substituting:

$m = \frac{-6-0}{0- 9} = \frac{-6}{-9} = \frac{2}{3}$

Line C:

Setting $y_{1} = x_{2}= 0, x_{1} = -12 , y_{2} = -8$ and substituting:

$m = \frac{-8-0}{0- (-12)} = \frac{-8}{12} = -\frac{2}{3}$

The product of the slopes of Line A and Line B is $- \frac{2}{3} \times \frac{2}{3} = -\frac{4}{9}$ , as is the product of the slopes of Line C and Line B; Since neither product is equal to , neither pair of lines is perpendicular. Also, the slopes of Line A and Line C are equal, so the lines are parallel, not perpendicular.

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Question

Lines P and Q are parallel. Find the value of .

Answer

Since these are complementary angles, we can set up the following equation.

Now we will use the quadratic formula to solve for .

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-10\pm \sqrt{10^2-4(1)(-180)}}{2(1)}$

$x=\frac{-10\pm \sqrt{100+720}}{2}$

$x=\frac{-10\pm \sqrt{820}}{2}$

$x=\frac{-10\pm \sqrt{4\cdot 205}}{2}$

$x=\frac{-10\pm 2\sqrt{ 205}}{2}$

$x=-5\pm \sqrt{ 205}$

Note, however, that the measure of an angle cannot be negative, so $x=-5- \sqrt{ 205}$ is not a viable answer. The correct answer, then, is $x=-5+ \sqrt{ 205}$

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Question

Which of the following lines is perpendicular to the line ?

Answer

Perpendicular lines will have slopes that are negative reciprocals of one another. Our first step will be to find the slope of the given line by putting the equation into slope-intercept form.

The slope of this line is . The negative reciprocal will be $+\frac{1}{3}$ , which will be the slope of the perpendicular line.

Now we need to find the answer choice with this slope by converting to slope-intercept form.

$y=\frac{4}{12}x-\frac{6}{12}$

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

This equation has a slope of $+\frac{1}{3}$ , and must be our answer.

Compare your answer with the correct one above

Question

Which set of lines is perpendicular?

Answer

Two lines are perpendicular to each other if their slopes are negative reciprocals. For example, if one line has a slope of 2, the line perpendicular to it has a slope of –1/2. One easy way to eliminate answer choices is to check if the slopes have the same sign, i.e. both positive or both negative. If so, they cannot be perpendicular. Several of the lines in the answer choices are of the form y = mx + b, where m is the slope and b is the y-intercept. We are only worried about the slope for the purposes of this question.

y = 3_x_ + 5 and y = 5_x_ + 3 both have positive slopes (m = 3 and m = 5, respectively), so they aren't perpendicular.

y = 3_x_/5 – 3 and y = 5_x_/3 + 3 both have positive slopes, so again they aren't perpendicular.

y = x – 1/2 and y = –x + 1/2 have slopes of m = 1 and m = –1, respectively. One is positive and one is negative, so that is a good sign. Let's take the negative reciprocal of 1. 1 /–1 = –1. So these two slopes are in fact negative reciprocals, and these two lines are perpendicular to each other. Even though we have found the correct answer, let's go through the other two choices to be sure.

The line between the points (1,3) and (3,5), and y = 4_x_ + 7: We need to find the slope of the first line. slope = rise / run = (_y_2 – _y_1) / (_x_2 – x_1) = (5 – 3) / (3 – 1) = 1. The slope of y = 4_x + 7 is also positive (m = 4), so the lines are not perpendicular.

The line between the points (7,4) and (4,7), and the line between the points (3,9) and (4,8): the first slope = (7 – 4) / (4 – 7) = –1 and the second slope = (8 – 9) / (4 – 3) = –1. They have the same slope, making them parallel, not perpendicular.

Compare your answer with the correct one above

Question

If two lines have slopes of $-5$ and $\frac{1}{5}$ , which statement about the lines is true?

Answer

Perpendicular lines have slopes that are the negative reciprocals of each other.

Compare your answer with the correct one above

Question

Which of the following lines is perpendicular to $y=3x-4$

Answer

The line which is perpendicular has a slope which is the negative inverse of the slope of the original line.

Compare your answer with the correct one above

Question

Which of the following equations represents a line that is perpendicular to the line with points and ?

Answer

If lines are perpendicular, then their slopes will be negative reciprocals.

First, we need to find the slope of the given line.

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{11-(-13)}{7-5}$

$m=\frac{24}{2}$

Because we know that our given line's slope is , the slope of the line perpendicular to it must be $-\frac{1}{12}$ .

Compare your answer with the correct one above

Question

Which of the following lines is perpindicular to

Answer

When determining if a two lines are perpindicular, we are only concerned about their slopes. Consider the basic equation of a line, , where m is the slope of the line. Two lines are perpindicular to each other if one slope is the negative and reciprocal of the other.

The first step of this problem is to get it into the form, , which is . Now we know that the slope, m, is . The reciprocal of that is $\frac{1}{3}$ , and the negative of that is $\frac{-1}{3}$ . Therefore, any line that has a slope of $\frac{-1}{3}$ will be perpindicular to the original line.

Compare your answer with the correct one above

Question

Which of the following lines is NOT perpendicular to ?

Answer

Perpendicular lines have slopes that are negative reciprocals. The given equation has a slope of ; therefore, any lime that is perpendicular to it will have a slope of $- \frac {1}{3}$ .

Compare your answer with the correct one above

Question

Line is given by $-\frac{6}{5}x+y=4$ .

Which of the following is perpendicular to ?

Answer

Putting the equation of the line into slope-intercept form, we get

$y=\frac{6}{5}x+4$

The slope of line , therefore, is $\frac{6}{5}$ .

In order for a line to be perpendicular to the given line, it must have a slope that is the negative reciprocal of line g's slope.

The slope of any given line perpendicular to line g must be $-\frac{5}{6}$ when written in slope-intercept form. In other words, the equation of the perpendicular line must be $y=-\frac{5}{6}x +k$ where k is any constant.

Written in standard form, the equation of this perpendicular line must be $\frac{5}{6}x+y=k.$

Therefore, the most appropriate answer is $\frac{5}{6}x+y=6.$

Compare your answer with the correct one above

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