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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 $\text{-} 1.$

$m_1\cdot m_2=\text{-} 1$

This graph shows one pair of perpendicular lines.

For the product of two slopes to equal $\text{-} 1,$ the slopes must be opposite reciprocals. For example, $m_1=\text{-} \frac{2}{3}$ and $m_2=\frac{3}{2}$ are opposite reciprocal slopes.
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Exercise

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Solution
Two lines are perpendicular if the product of their slopes is $\text{-} 1.$ Notice that the equations are given in slope-intercept form. Thus, we can see that the lines have the slopes $m_1=\text{-} 2 \quad \text{and} \quad m_2=\frac{2}{5}.$ Let's multiply the slopes to determine if their product is $\text{-} 1.$
$m_1 \cdot m_2 = \text{-} 1$
${\color{#0000FF}{\text{-} 2}} \cdot {\color{#009600}{\dfrac{2}{5}}} \stackrel{?}{=} \text{-} 1$
$\dfrac{\text{-} 2 \cdot 2}{5} \stackrel{?}{=} \text{-} 1$
$\dfrac{\text{-} 4}{5} \neq \text{-} 1$
The product of the slopes does not equal $\text{-}1.$ Therefore, the lines are not perpendicular.
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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 $(0,\text{-} 1).$

Show Solution
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 $\text{-} 1.$ $m_1 \cdot m_2 = \text{-}1$ The given line has the slope $m_1=\frac{2}{3}.$ Thus, we can use the product to determine $m_2.$
$m_1\cdot m_2=\text{-}1$
${\color{#0000FF}{\dfrac{2}{3}}}\cdot m_2=\text{-}1$
$2\cdot m_2=\text{-}3$
$m_2=\text{-} \dfrac{3}{2}$
The slope of the other line is $m=\text{-} \frac{3}{2},$ so we can write the incomplete equation of this line as $y=\text{-} \dfrac{3}{2}x+b.$ Recall that the line must pass through the point $(0,\text{-} 1).$ To find the $y$-intercept, we can substitute the point $(0,\text{-} 1)$ for $x$ and $y$ in the incomplete equation above.
$y=\text{-} \dfrac{3}{2}x+b$
${\color{#009600}{\text{-} 1}}=\text{-} \dfrac{3}{2} \cdot{\color{#0000FF}{0}}+b$
$\text{-} 1=b$
$b=\text{-} 1$
The $y$-intercept of the line is $b=\text{-} 1.$ This gives the equation $y=\text{-} \dfrac{3}{2}x - 1.$