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SAT Math : Lines

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Example Questions

Example Question #2 : How To Find Out If Lines Are Perpendicular

Which of the following lines is perpindicular to y-3x=7

Possible Answers:

None of the other answers

$y=\frac{-1}{3}x - 4$

$y = \frac{1}{3}x +7$

y=-3x-2

$y-\frac{1}{3}x=12$

Correct answer:

$y=\frac{-1}{3}x - 4$

Explanation:

When determining if a two lines are perpindicular, we are only concerned about their slopes. Consider the basic equation of a line, y=mx+b , 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, y=mx+b , which is y=3x+7 . 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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Example Question #2 : How To Find Out If Lines Are Perpendicular

Which of the following equations represents a line that is perpendicular to the line with points (5,-13) and (7,11) ?

Possible Answers:

y=-x+4

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

y=-12x+6

$y=-\frac{1}{12}x+4$

y=12x-4

Correct answer:

$y=-\frac{1}{12}x+4$

Explanation:

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

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

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

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

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

m=12

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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Example Question #1 : How To Find Out If Lines Are Perpendicular

Which of the following lines is NOT perpendicular to y=3x-3 ?

Possible Answers:

3y= -x + 2

2x = -6y + 4

.5x + 1.5y = 10.5

12y = -4x -15

x=y+3

Correct answer:

x=y+3

Explanation:

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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Example Question #1 : How To Find Out If Lines Are Perpendicular

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

Which of the following is perpendicular to ?

Possible Answers:

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

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

$\frac{6}{5}x-y=6$

$\frac{6}{5}x+y=3$

$\frac{3}{5}x+y=-2$

Correct answer:

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

Explanation:

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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Example Question #2 : How To Find Out If Lines Are Perpendicular

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

The equation of Line A is y = 3x+ 17 .

The equation of Line B is -9x + 3y =17 .

The equation of Line is 6x - 2y = 17 .

Which of the following is a true statement?

Possible Answers:

Line A and Line C are perpendicular to each other, but Line B is perpendicular to neither Line A nor Line C.

None of the statements in the other choices is true.

Line A and Line B are perpendicular to each other, but Line C is perpendicular to neither Line A nor Line B.

No two of Line A, Line B, and Line C are perpendicular to each other.

Line B and Line C are perpendicular to each other, but Line A is perpendicular to neither Line B nor Line C.

Correct answer:

No two of Line A, Line B, and Line C are perpendicular to each other.

Explanation:

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 y = mx+ b , where is the slope of the line.

Line A:

y = 3x+ 17 is already in this form. The slope of Line A is the coefficient of , which is 3.

Line B:

-9x + 3y =17

Isolate by working the same operations on both sides:

-9x + 3y + 9x =17 + 9x

3y = 9x + 17

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

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

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

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

Line C:

6x - 2y = 17

6x - 2y - 6x= 17 - 6x

- 2y = - 6x + 17

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

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

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

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 $3 \times 3 = 9 \ne -1$ . Therefore, no two of the lines are perpendicular.

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Example Question #1 : How To Find Out If Lines Are Perpendicular

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

The equation of Line A is 2x+ 7y = 13 .

The equation of Line B is 7x - 2y = 13 .

The equation of Line C is 2x-7y = 13 .

Which of the following is a true statement?

Possible Answers:

Line A and Line C are perpendicular to each other, but Line B is perpendicular to neither Line A nor Line C.

Line A and Line B are perpendicular to each other, but Line C is perpendicular to neither Line A nor Line B.

Line B and Line C are perpendicular to each other, but Line A is perpendicular to neither Line B nor Line C.

None of the statements in the other choices is true.

No two of Line A, Line B, and Line C are perpendicular to each other.

Correct answer:

Line A and Line B are perpendicular to each other, but Line C is perpendicular to neither Line A nor Line B.

Explanation:

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 y = mx+ b , where is the slope of the line.

Line A:

2x+ 7y = 13

Isolate by working the same operations on both sides:

2x+ 7y - 2x = 13 - 2x

7y = - 2x+ 13

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

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

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

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:

7x - 2y = 13

7x - 2y -7x = 13-7x

- 2y = -7x + 13

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

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

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

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

Line C:

2x-7y = 13

2x-7y - 2x = 13 - 2x

-7y = - 2x + 13

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

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

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

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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Example Question #7 : How To Find Out If Lines Are Perpendicular

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

Line A has intercepts (0,4) and (6,0 ) .

Line B has intercepts (0, -6) and (9, 0) .

Line C has intercepts (0,- 8 ) and (-12, 0 ) .

Which of the following statements is true?

Possible Answers:

No two of Line A, Line B, and Line C are perpendicular to each other.

Line A and Line B are perpendicular to each other, but Line C is perpendicular to neither Line A nor Line B.

Line B and Line C are perpendicular to each other, but Line A is perpendicular to neither Line B nor Line C.

Line A and Line C are perpendicular to each other, but Line B is perpendicular to neither Line A nor Line C.

None of the statements in the other choices is true.

Correct answer:

No two of Line A, Line B, and Line C are perpendicular to each other.

Explanation:

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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Example Question #1 : How To Find Out If Lines Are Perpendicular

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

Possible Answers:

$y=-3x-4$

$y=\frac{1}{3}x-4$

$y=\frac{1}{3}x+4$

$y=-\frac{1}{3}x-4$

Correct answer:

$y=-\frac{1}{3}x-4$

Explanation:

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

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Example Question #81 : Lines

Trans

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

Possible Answers:

$x=\pm \sqrt{ 205}$

$x=5 + \sqrt{ 205}$

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

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

$x= \sqrt{ 205}$

Correct answer:

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

Explanation:

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

x^2+4+10x-4=180

x^2+10x=180

x^2+10x-180=0

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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Example Question #1 : Other Lines

In the xy -plane, line l is given by the equation 2x - 3y = 5. If line l passes through the point (a ,1), what is the value of a ?

Possible Answers:

-2

-1

Correct answer: 4

Explanation:

The equation of line l relates x -values and y -values that lie along the line. The question is asking for the x -value of a point on the line whose y -value is 1, so we are looking for the x -value on the line when the y-value is 1. In the equation of the line, plug 1 in for y and solve for x:

2x - 3(1) = 5

2x - 3 = 5

2x = 8

x = 4. So the missing x-value on line l is 4.

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SAT Math : Lines

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #2 : How To Find Out If Lines Are Perpendicular

Example Question #2 : How To Find Out If Lines Are Perpendicular

Example Question #1 : How To Find Out If Lines Are Perpendicular

Example Question #1 : How To Find Out If Lines Are Perpendicular

Example Question #2 : How To Find Out If Lines Are Perpendicular

Example Question #1 : How To Find Out If Lines Are Perpendicular

Example Question #7 : How To Find Out If Lines Are Perpendicular

Example Question #1 : How To Find Out If Lines Are Perpendicular

Example Question #81 : Lines

Example Question #1 : Other Lines

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