Find the Perpendicular Line Passing through (3,-4) and perpendicular to the line whose equation is y=1/6x+1

Passing through $(3, - 4)$ and perpendicular to the line whose equation is $y = \frac{1}{6} x + 1$

Step 1

Find the slope when $y = \frac{1}{6} x + 1$ .

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.

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

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

Reorder terms.

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

Using the slope-intercept form, the slope is $\frac{1}{6}$ .

$m = \frac{1}{6}$

Step 2

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

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

Step 3

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

Multiply the numerator by the reciprocal of the denominator.

$m_{perpendicular} = - (1 \cdot 6)$

Multiply $- (1 \cdot 6)$ .

Multiply $6$ by $1$ .

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

Multiply $- 1$ by $6$ .

$m_{perpendicular} = - 6$

Step 4

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

Use the slope $- 6$ and a given point $(3, - 4)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (- 4) = - 6 \cdot (x - (3))$

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

$y + 4 = - 6 \cdot (x - 3)$

Step 5

Solve for $y$ .

Simplify $- 6 \cdot (x - 3)$ .

Rewrite.

$y + 4 = 0 + 0 - 6 \cdot (x - 3)$

Simplify by adding zeros.

$y + 4 = - 6 \cdot (x - 3)$

Apply the distributive property.

$y + 4 = - 6 x - 6 \cdot - 3$

Multiply $- 6$ by $- 3$ .

$y + 4 = - 6 x + 18$

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

Subtract $4$ from both sides of the equation.

$y = - 6 x + 18 - 4$

Subtract $4$ from $18$ .

$y = - 6 x + 14$

Step 6

Find the Perpendicular Line Passing through (3,-4) and perpendicular to the line whose equation is y=1/6x+1

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