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High School Math : How to find the equation of a perpendicular line

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Example Questions

Example Question #1370 : High School Math

Write an equation in slope-intercept form for the line that passes through $\left ( 3,1 \right )$ and that is perpendicular to a line which passes through the two points $\left ( 3,1 \right )$ and $\left ( -2,5 \right )$ .

Possible Answers:

$y=\frac{5}{4}X+\frac{11}{4}$

y=5X-4

$y=\frac{5}{4}X-\frac{11}{4}$

$y=-\frac{4}{5}X+11$

Correct answer:

$y=\frac{5}{4}X-\frac{11}{4}$

Explanation:

Find the slope of the line through the two points. It is $-\frac{4}{5}$ .

Since the slope of a perpendicular line is the negative reciprocal of the original line, the new line's slope is $\frac{5}{4}$ . Plug the slope and one of the points into the point-slope formula $y-y_{1}=m(x-x_{1})$ . Isolate for .

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Example Question #1 : How To Find The Equation Of A Perpendicular Line

Find the equation of a line perpendicular to y=-3x+10

Possible Answers:

y=-3X+10

y=-3X-5

y=3X+2

$y=\frac{1}{3}X+8$

Correct answer:

$y=\frac{1}{3}X+8$

Explanation:

Since a perpendicular line has a slope that is the negative reciprocal of the original line, the new slope is $\frac{1}{3}$ . There is only one answer with the correct slope.

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Example Question #2 : Perpendicular Lines

Find the equation (in slope-intercept form) of a line perpendicular to 3X+2Y=11 .

Possible Answers:

$y=3X+\frac{11}{2}$

$y=-\frac{3}{2}X+11$

$y=\frac{2}{3}X+\frac{11}{2}$

$y=2X+\frac{11}{2}$

Correct answer:

$y=\frac{2}{3}X+\frac{11}{2}$

Explanation:

First, find the slope of the original line, which is $-\frac{3}{2}$ . You can do this by isolating for so that the equation is in slope-intercept form. Once you find the slope, just replace the in the original equation withe the negative reciprocal (perpendicular lines have a negative reciprocal slope for each other). Thus, your answer is

$y=\frac{2}{3}X+\frac{11}{2}$

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Example Question #3 : Perpendicular Lines

Given the equation 2x+4y=-12 and the point (1,2) , find the equation of a line that is perpendicular to the original line and passes through the given point.

Possible Answers:

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

y=2x-3

y=2x-12

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

y=2x

Correct answer:

y=2x

Explanation:

In order for two lines to be perpendicular, their slopes must be opposites and recipricals of each other. The first step is to find the slope of the given equation:

2x+4y=-12

4y=-2x-12

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

Therefore, the slope of the perpendicular line must be . Using the point-slope formula, we can find the equation of the new line:

y-2=2(x-1)

y-2=2x-2

y=2x

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Example Question #4 : Perpendicular Lines

What line is perpendicular to $y=\frac{1}{2}x+3$ through (6,5) ?

Possible Answers:

y+3x=15

y+6x=7

y-3x=21

y-2x=21

y+2x=17

Correct answer:

y+2x=17

Explanation:

Perdendicular lines have slopes that are opposite reciprocals. The slope of the old line is $\frac{1}{2}$ , so the new slope is .

Plug the new slope and the given point into the slope intercept equation to calculate the intercept:

y=mx+b or 5=-2(6)+b , so b=17 .

Thus y=-2x+17 , or y+2x=17 .

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Example Question #5 : Perpendicular Lines

What is the equation, in slope-intercept form, of the perpendicular bisector of the line segment that connects the points $\small (2,2)$ and $\small (8,6)$ ?

Possible Answers:

y=-3x+23

$y=\frac{3}{2}x+\frac{23}{2}$

$y=-\frac{3}{2}x+\frac{7}{2}$

$y=-\frac{3}{2}x+\frac{23}{2}$

$y=-\frac{3}{2}x-\frac{7}{2}$

Correct answer:

$y=-\frac{3}{2}x+\frac{23}{2}$

Explanation:

First, calculate the slope of the line segment between the given points.

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{8-2}=\frac{4}{6}=\frac{2}{3}$

We want a line that is perpendicular to this segment and passes through its midpoint. The slope of a perpendicular line is the negative inverse. The slope of the perpendicular bisector will be $m=-\frac{3}{2}$ .

Next, we need to find the midpoint of the segment, using the midpoint formula.

$mid=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{x})=(\frac{2+8}{2},\frac{2+6}{2})=(5,4)$

Using the midpoint and the slope, we can solve for the value of the y-intercept.

y=mx+b

$4=(-\frac{3}{2})(5)+b$

$4=\frac{8}{2}=-\frac{15}{2}+b$

$\frac{23}{2}=b$

Using this value, we can write the equation for the perpendicular bisector in slope-intercept form.

$y=-\frac{3}{2}x+\frac{23}{2}$

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Example Question #6 : Perpendicular Lines

What line is perpendicular to y=2x-3 through (6,-2) ?

Possible Answers:

4x+3y=-1

-2x+y=6

x+2y=2

2x-y=10

3x-2y=5

Correct answer:

x+2y=2

Explanation:

The equation is given in the slope-intercept form, so we know the slope is . To have perpendicular lines, the new slope must be the opposite reciprocal of the old slope, or $\frac{-1}{2}$

Then plug the new slope and the point into the slope-intercept form of the equation:

$-2=\frac{-1}{2}(6)+b$ so -2=-3+b so b=1

So the new equation becomes: $y=\frac{-1}{2}x+1$ and in standard form x+2y=2

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High School Math : How to find the equation of a perpendicular line

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1370 : High School Math

Example Question #1 : How To Find The Equation Of A Perpendicular Line

Example Question #2 : Perpendicular Lines

Example Question #3 : Perpendicular Lines

Example Question #4 : Perpendicular Lines

Example Question #5 : Perpendicular Lines

Example Question #6 : Perpendicular Lines

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