Find Any Equation Perpendicular to the Line y=-1/3x+1

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

Choose a point that the perpendicular line will pass through.

$(0, 0)$

Find the slope when $y = \frac{- 1}{3} x + 1$ .

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.

$y = m x + b$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

Combine $x$ and $\frac{1}{3}$ .

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

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

Reorder terms.

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

Remove parentheses.

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

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

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

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

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

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 $3$ by $1$ .

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

Multiply $- 1$ by $3$ .

$m_{perpendicular} = 3$

Multiply $- 1$ by $- 3$ .

$m_{perpendicular} = 3$

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

Use the slope $3$ and a given point $(0, 0)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (0) = 3 \cdot (x - (0))$

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

$y + 0 = 3 \cdot (x + 0)$

Solve for $y$ .

Add $y$ and $0$ .

$y = 3 \cdot (x + 0)$

Add $x$ and $0$ .

$y = 3 x$

Find Any Equation Perpendicular to the Line y=-1/3x+1

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