Find the Slope of a Perpendicular Line 3x+5y=1

$3 x + 5 y = 1$

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.

$y = m x + b$

Subtract $3 x$ from both sides of the equation.

$5 y = 1 - 3 x$

Divide each term in $5 y = 1 - 3 x$ by $5$ and simplify.

Divide each term in $5 y = 1 - 3 x$ by $5$ .

$\frac{5 y}{5} = \frac{1}{5} + \frac{- 3 x}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 y}{5} = \frac{1}{5} + \frac{- 3 x}{5}$

Divide $y$ by $1$ .

$y = \frac{1}{5} + \frac{- 3 x}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$y = \frac{1}{5} - \frac{3 x}{5}$

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

Reorder $\frac{1}{5}$ and $- \frac{3 x}{5}$ .

$y = - \frac{3 x}{5} + \frac{1}{5}$

Reorder terms.

$y = - (\frac{3}{5} x) + \frac{1}{5}$

Remove parentheses.

$y = - \frac{3}{5} x + \frac{1}{5}$

Using the slope-intercept form, the slope is $- \frac{3}{5}$ .

$m = - \frac{3}{5}$

The equation of a perpendicular line to $y = - \frac{3}{5} x + \frac{1}{5}$ must have a slope that is the negative reciprocal of the original slope.

$m_{perpendicular} = - \frac{1}{- \frac{3}{5}}$

Simplify the result.

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

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

$m_{perpendicular} = - \frac{- 1 \cdot - 1}{- \frac{3}{5}}$

Move the negative in front of the fraction.

$m_{perpendicular} = \frac{1}{\frac{3}{5}}$

Multiply the numerator by the reciprocal of the denominator.

$m_{perpendicular} = 1 (\frac{5}{3})$

Multiply $\frac{5}{3}$ by $1$ .

$m_{perpendicular} = \frac{5}{3}$

Multiply $- - \frac{5}{3}$ .

Multiply $- 1$ by $- 1$ .

$m_{perpendicular} = 1 (\frac{5}{3})$

Multiply $\frac{5}{3}$ by $1$ .

$m_{perpendicular} = \frac{5}{3}$