Writing Equations of Perpendicular Lines

Linear functions are a family of functions that all have a constant rate of change. There are additional ways that pairs of lines can relate. One such way is if the lines are perpendicular. In this section, what makes two lines perpendicular and the ways in which their rules and graphs are similar/different will be explored.

Rule

Perpendicular Lines

Lines that intersect at right angles are called perpendicular lines. All vertical lines are perpendicular to all horizontal lines. Lines are perpendicular if and only if the product of their slopes,

m_{1}

and

m_{2},

is

- 1 .

$m_{1} \cdot m_{2} = - 1$

This graph shows one pair of perpendicular lines.

For the product of two slopes to equal $- 1,$ the slopes must be opposite reciprocals. For example, $m_{1} = - \frac{2}{3}$ and $m_{2} = \frac{3}{2}$ are opposite reciprocal slopes.

Are the lines perpendicular?

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Exercise

Are the lines shown in the graph perpendicular? Justify your answer.

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Solution

Two lines are perpendicular if the product of their slopes is

- 1 .

Notice that the equations are given in slope-intercept form. Thus, we can see that the lines have the slopes

m_{1} = - 2 and m_{2} = \frac{2}{5} .

Let's multiply the slopes to determine if their product is

- 1 .

m_{1} \cdot m_{2} = - 1

SubstituteII

$m_{1} = - 2$ , $m_{2} = \frac{2}{5}$

- 2 \cdot \frac{2}{5} = ? - 1

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

\frac{- 2 \cdot 2}{5} = ? - 1

Multiply

\frac{- 4}{5} \neq = - 1

The product of the slopes does not equal

- 1 .

Therefore, the lines are not perpendicular.

Find the perpendicular line

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Exercise

Write the equation of the line that's perpendicular to $y = \frac{2}{3} x - 1$ and passes through the point $(0, - 1) .$

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Solution

To write the equation of the line in slope-intercept form, we need its slope and its

y

-intercept. To ensure the lines are perpendicular, their slopes,

m_{1}

and

m_{2},

must multiply to equal

- 1 .

m_{1} \cdot m_{2} = - 1

The given line has the slope

m_{1} = \frac{2}{3} .

Thus, we can use the product to determine

m_{2} .

m_{1} \cdot m_{2} = - 1

Substitute

$m_{1} = \frac{2}{3}$

\frac{2}{3} \cdot m_{2} = - 1

MultEqn

$LHS \cdot 3 = RHS \cdot 3$

2 \cdot m_{2} = - 3

DivEqn

$LHS / 2 = RHS / 2$

m_{2} = - \frac{3}{2}

The slope of the other line is

m = - \frac{3}{2},

so we can write the incomplete equation of this line as

y = - \frac{3}{2} x + b .

Recall that the line must pass through the point

(0, - 1) .

To find the

y

-intercept, we can substitute the point

(0, - 1)

for

x

and

y

in the incomplete equation above.

y = - \frac{3}{2} x + b

SubstituteII

$x = 0$ , $y = - 1$

- 1 = - \frac{3}{2} \cdot 0 + b

ZeroPropMult

Zero Property of Multiplication

- 1 = b

RearrangeEqn

Rearrange equation

b = - 1

The

y

-intercept of the line is

b = - 1 .

This gives the equation

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