Find Any Equation Perpendicular to the Line y=-8x+3

$y = - 8 x + 3$

Step 1

Choose a point that the perpendicular line will pass through.

$(0, 0)$

Step 2

Use the slope-intercept form to find the slope.

The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

Using the slope-intercept form, the slope is $- 8$ .

$m = - 8$

Step 3

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

$m_{perpendicular} = - \frac{1}{- 8}$

Step 4

Simplify $- \frac{1}{- 8}$ to find the slope of the perpendicular line.

Move the negative in front of the fraction.

$m_{perpendicular} = \frac{1}{8}$

Multiply $- - \frac{1}{8}$ .

Multiply $- 1$ by $- 1$ .

$m_{perpendicular} = 1 (\frac{1}{8})$

Multiply $\frac{1}{8}$ by $1$ .

$m_{perpendicular} = \frac{1}{8}$

Step 5

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

Use the slope $\frac{1}{8}$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (0) = \frac{1}{8} \cdot (x - (0))$

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

$y + 0 = \frac{1}{8} \cdot (x + 0)$

Step 6

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

Solve for $y$ .

Add $y$ and $0$ .

$y = \frac{1}{8} \cdot (x + 0)$

Simplify $\frac{1}{8} \cdot (x + 0)$ .

Add $x$ and $0$ .

$y = \frac{1}{8} \cdot x$

Combine $\frac{1}{8}$ and $x$ .

$y = \frac{x}{8}$

Reorder terms.

$y = \frac{1}{8} x$

Step 7

Find Any Equation Perpendicular to the Line y=-8x+3

Related Questions

Support LCMGCF.com by disabling your adblocker.