Find the Perpendicular Line 4x+3y=9 and passes through (-2,3)

$4 x + 3 y = 9$ and passes through $(- 2, 3)$

Solve $4 x + 3 y = 9$ .

Subtract $4 x$ from both sides of the equation.

$3 y = 9 - 4 x$

Divide each term in $3 y = 9 - 4 x$ by $3$ and simplify.

Divide each term in $3 y = 9 - 4 x$ by $3$ .

$\frac{3 y}{3} = \frac{9}{3} + \frac{- 4 x}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{9}{3} + \frac{- 4 x}{3}$

Divide $y$ by $1$ .

$y = \frac{9}{3} + \frac{- 4 x}{3}$

Simplify the right side.

Simplify each term.

Divide $9$ by $3$ .

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

Move the negative in front of the fraction.

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

Find the slope when $y = 3 - \frac{4 x}{3}$ .

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 $3$ and $- \frac{4 x}{3}$ .

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

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

Reorder terms.

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

Remove parentheses.

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

Using the slope-intercept form, the slope is $- \frac{4}{3}$ .

$m = - \frac{4}{3}$

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

$m_{perpendicular} = - \frac{1}{- \frac{4}{3}}$

Simplify $- \frac{1}{- \frac{4}{3}}$ to find the slope of the perpendicular line.

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

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

$m_{perpendicular} = - \frac{- 1 \cdot - 1}{- \frac{4}{3}}$

Move the negative in front of the fraction.

$m_{perpendicular} = \frac{1}{\frac{4}{3}}$

Multiply the numerator by the reciprocal of the denominator.

$m_{perpendicular} = 1 (\frac{3}{4})$

Multiply $\frac{3}{4}$ by $1$ .

$m_{perpendicular} = \frac{3}{4}$

Multiply $- - \frac{3}{4}$ .

Multiply $- 1$ by $- 1$ .

$m_{perpendicular} = 1 (\frac{3}{4})$

Multiply $\frac{3}{4}$ by $1$ .

$m_{perpendicular} = \frac{3}{4}$

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

Use the slope $\frac{3}{4}$ and a given point $(- 2, 3)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

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

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

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

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

Solve for $y$ .

Simplify $\frac{3}{4} \cdot (x + 2)$ .

Rewrite.

$y - 3 = 0 + 0 + \frac{3}{4} \cdot (x + 2)$

Simplify by adding zeros.

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

Apply the distributive property.

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

Combine $\frac{3}{4}$ and $x$ .

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

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Add $3$ to both sides of the equation.

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

To write $3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $3$ by $2$ .

$y = \frac{3 x}{4} + \frac{3 + 6}{2}$

Add $3$ and $6$ .

$y = \frac{3 x}{4} + \frac{9}{2}$

Reorder terms.

$y = \frac{3}{4} x + \frac{9}{2}$

Find the Perpendicular Line 4x+3y=9 and passes through (-2,3)

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