Find the Perpendicular Line What is the equation of the line that is perpendicular to the line defined by the equation 2y=3x+7 and goes through the point (3,2) ?

What is the equation of the line that is perpendicular to the line defined by the equation $2 y = 3 x + 7$ and goes through the point $(3, 2)$ ?

Divide each term in $2 y = 3 x + 7$ by $2$ and simplify.

Divide each term in $2 y = 3 x + 7$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Find the slope when $y = \frac{3 x}{2} + \frac{7}{2}$ .

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 terms.

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

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

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

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

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

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

Multiply the numerator by the reciprocal of the denominator.

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

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

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

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

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

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

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

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

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

Solve for $y$ .

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

Rewrite.

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

Simplify terms.

Apply the distributive property.

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

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

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

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{2}{3}$ into the numerator.

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

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 2$ by $- 1$ .

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

Move $2$ to the left of $x$ .

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

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

Add $2$ to both sides of the equation.

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

Add $2$ and $2$ .

$y = - \frac{2 x}{3} + 4$

Reorder terms.

$y = - (\frac{2}{3} x) + 4$

Remove parentheses.

$y = - \frac{2}{3} x + 4$

Find the Perpendicular Line What is the equation of the line that is perpendicular to the line defined by the equation 2y=3x+7 and goes through the point (3,2) ?

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