Find the Perpendicular Line What is an equation of the line that passes through the point (6,-2) and is perpendicular to the line 6x+y=2 ?

What is an equation of the line that passes through the point $(6, - 2)$ and is perpendicular to the line $6 x + y = 2$ ?

Step 1

Write the problem as a mathematical expression.

$(6, - 2)$ , $6 x + y = 2$

Step 2

Subtract $6 x$ from both sides of the equation.

$y = 2 - 6 x$

Step 3

Find the slope when $y = 2 - 6 x$ .

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 $2$ and $- 6 x$ .

$y = - 6 x + 2$

Using the slope-intercept form, the slope is $- 6$ .

$m = - 6$

Step 4

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

$m_{perpendicular} = - \frac{1}{- 6}$

Step 5

Simplify $- \frac{1}{- 6}$ to find the slope of the perpendicular line.

Move the negative in front of the fraction.

$m_{perpendicular} = \frac{1}{6}$

Multiply $- - \frac{1}{6}$ .

Multiply $- 1$ by $- 1$ .

$m_{perpendicular} = 1 (\frac{1}{6})$

Multiply $\frac{1}{6}$ by $1$ .

$m_{perpendicular} = \frac{1}{6}$

Step 6

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

Use the slope $\frac{1}{6}$ and a given point $(6, - 2)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (- 2) = \frac{1}{6} \cdot (x - (6))$

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

$y + 2 = \frac{1}{6} \cdot (x - 6)$

Step 7

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

Solve for $y$ .

Simplify $\frac{1}{6} \cdot (x - 6)$ .

Rewrite.

$y + 2 = 0 + 0 + \frac{1}{6} \cdot (x - 6)$

Simplify by adding zeros.

$y + 2 = \frac{1}{6} \cdot (x - 6)$

Apply the distributive property.

$y + 2 = \frac{1}{6} x + \frac{1}{6} \cdot - 6$

Combine $\frac{1}{6}$ and $x$ .

$y + 2 = \frac{x}{6} + \frac{1}{6} \cdot - 6$

Cancel the common factor of $6$ .

Factor $6$ out of $- 6$ .

$y + 2 = \frac{x}{6} + \frac{1}{6} \cdot (6 (- 1))$

Cancel the common factor.

$y + 2 = \frac{x}{6} + \frac{1}{6} \cdot (6 \cdot - 1)$

Rewrite the expression.

$y + 2 = \frac{x}{6} - 1$

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

Subtract $2$ from both sides of the equation.

$y = \frac{x}{6} - 1 - 2$

Subtract $2$ from $- 1$ .

$y = \frac{x}{6} - 3$

Reorder terms.

$y = \frac{1}{6} x - 3$

Step 8

Find the Perpendicular Line What is an equation of the line that passes through the point (6,-2) and is perpendicular to the line 6x+y=2 ?

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