Find the Perpendicular Line What is an equation of the line that passes through the point (1,6) and is perpendicular to the line x+3y=27 ?

What is an equation of the line that passes through the point $(1, 6)$ and is perpendicular to the line $x + 3 y = 27$ ?

Step 1

Solve $x + 3 y = 27$ .

Subtract $x$ from both sides of the equation.

$3 y = 27 - x$

Divide each term in $3 y = 27 - x$ by $3$ and simplify.

Divide each term in $3 y = 27 - x$ by $3$ .

$\frac{3 y}{3} = \frac{27}{3} + \frac{- x}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{27}{3} + \frac{- x}{3}$

Divide $y$ by $1$ .

$y = \frac{27}{3} + \frac{- x}{3}$

Simplify the right side.

Simplify each term.

Divide $27$ by $3$ .

$y = 9 + \frac{- x}{3}$

Move the negative in front of the fraction.

$y = 9 - \frac{x}{3}$

Step 2

Find the slope when $y = 9 - \frac{x}{3}$ .

Rewrite in slope-intercept form.

The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

Reorder $9$ and $- \frac{x}{3}$ .

$y = - \frac{x}{3} + 9$

Write in $y = m x + b$ form.

Reorder terms.

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

Remove parentheses.

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

Using the slope-intercept form, the slope is $- \frac{1}{3}$ .

$m = - \frac{1}{3}$

Step 3

The equation of a perpendicular line must have a slope that is the negative reciprocal of the original slope.

$m_{perpendicular} = - \frac{1}{- \frac{1}{3}}$

Step 4

Simplify $- \frac{1}{- \frac{1}{3}}$ to find the slope of the perpendicular line.

Cancel the common factor of $1$ and $- 1$ .

Rewrite $1$ as $- 1 (- 1)$ .

$m_{perpendicular} = - \frac{- 1 \cdot - 1}{- \frac{1}{3}}$

Move the negative in front of the fraction.

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

Multiply the numerator by the reciprocal of the denominator.

$m_{perpendicular} = 1 \cdot 3$

Multiply $- - (1 \cdot 3)$ .

Multiply $3$ by $1$ .

$m_{perpendicular} = - (- 1 \cdot 3)$

Multiply $- 1$ by $3$ .

$m_{perpendicular} = 3$

Multiply $- 1$ by $- 3$ .

$m_{perpendicular} = 3$

Step 5

Find the equation of the perpendicular line using the point-slope formula.

Use the slope $3$ and a given point $(1, 6)$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (6) = 3 \cdot (x - (1))$

Simplify the equation and keep it in point-slope form.

$y - 6 = 3 \cdot (x - 1)$

Step 6

Solve for $y$ .

Simplify $3 \cdot (x - 1)$ .

Rewrite.

$y - 6 = 0 + 0 + 3 \cdot (x - 1)$

Simplify by adding zeros.

$y - 6 = 3 \cdot (x - 1)$

Apply the distributive property.

$y - 6 = 3 x + 3 \cdot - 1$

Multiply $3$ by $- 1$ .

$y - 6 = 3 x - 3$

Move all terms not containing $y$ to the right side of the equation.

Add $6$ to both sides of the equation.

$y = 3 x - 3 + 6$

Add $- 3$ and $6$ .

$y = 3 x + 3$

Step 7

Find the Perpendicular Line What is an equation of the line that passes through the point (1,6) and is perpendicular to the line x+3y=27 ?

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