Find the Perpendicular Line What is an equation of the line that passes through the point (6,1) and is perpendicular to the line 2x+3y=18 ?

What is an equation of the line that passes through the point $(6, 1)$ and is perpendicular to the line $2 x + 3 y = 18$ ?

Step 1

Write the problem as a mathematical expression.

$(6, 1)$ , $2 x + 3 y = 18$

Step 2

Solve $2 x + 3 y = 18$ .

Subtract $2 x$ from both sides of the equation.

$3 y = 18 - 2 x$

Divide each term in $3 y = 18 - 2 x$ by $3$ and simplify.

Divide each term in $3 y = 18 - 2 x$ by $3$ .

$\frac{3 y}{3} = \frac{18}{3} + \frac{- 2 x}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{18}{3} + \frac{- 2 x}{3}$

Divide $y$ by $1$ .

$y = \frac{18}{3} + \frac{- 2 x}{3}$

Simplify the right side.

Simplify each term.

Divide $18$ by $3$ .

$y = 6 + \frac{- 2 x}{3}$

Move the negative in front of the fraction.

$y = 6 - \frac{2 x}{3}$

Step 3

Find the slope when $y = 6 - \frac{2 x}{3}$ .

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 $6$ and $- \frac{2 x}{3}$ .

$y = - \frac{2 x}{3} + 6$

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

Reorder terms.

$y = - (\frac{2}{3} x) + 6$

Remove parentheses.

$y = - \frac{2}{3} x + 6$

Using the slope-intercept form, the slope is $- \frac{2}{3}$ .

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

Step 4

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

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

Step 5

Simplify $- \frac{1}{- \frac{2}{3}}$ to find the slope of the perpendicular line.

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

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

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

Move the negative in front of the fraction.

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

Multiply the numerator by the reciprocal of the denominator.

$m_{perpendicular} = 1 (\frac{3}{2})$

Multiply $\frac{3}{2}$ by $1$ .

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

Multiply $- - \frac{3}{2}$ .

Multiply $- 1$ by $- 1$ .

$m_{perpendicular} = 1 (\frac{3}{2})$

Multiply $\frac{3}{2}$ by $1$ .

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

Step 6

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

Use the slope $\frac{3}{2}$ and a given point $(6, 1)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (1) = \frac{3}{2} \cdot (x - (6))$

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

$y - 1 = \frac{3}{2} \cdot (x - 6)$

Step 7

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

Solve for $y$ .

Simplify $\frac{3}{2} \cdot (x - 6)$ .

Rewrite.

$y - 1 = 0 + 0 + \frac{3}{2} \cdot (x - 6)$

Simplify by adding zeros.

$y - 1 = \frac{3}{2} \cdot (x - 6)$

Apply the distributive property.

$y - 1 = \frac{3}{2} x + \frac{3}{2} \cdot - 6$

Combine $\frac{3}{2}$ and $x$ .

$y - 1 = \frac{3 x}{2} + \frac{3}{2} \cdot - 6$

Cancel the common factor of $2$ .

Factor $2$ out of $- 6$ .

$y - 1 = \frac{3 x}{2} + \frac{3}{2} \cdot (2 (- 3))$

Cancel the common factor.

$y - 1 = \frac{3 x}{2} + \frac{3}{2} \cdot (2 \cdot - 3)$

Rewrite the expression.

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

Multiply $3$ by $- 3$ .

$y - 1 = \frac{3 x}{2} - 9$

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

Add $1$ to both sides of the equation.

$y = \frac{3 x}{2} - 9 + 1$

Add $- 9$ and $1$ .

$y = \frac{3 x}{2} - 8$

Reorder terms.

$y = \frac{3}{2} x - 8$

Step 8

Find the Perpendicular Line What is an equation of the line that passes through the point (6,1) and is perpendicular to the line 2x+3y=18 ?

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