Find the Perpendicular Line y=2x+1 at (12,5)

$y = 2 x + 1$ at $(12, 5)$

Step 1

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 $2$ .

$m = 2$

Step 2

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

$m_{perpendicular} = - \frac{1}{2}$

Step 3

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

Use the slope $- \frac{1}{2}$ and a given point $(12, 5)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (5) = - \frac{1}{2} \cdot (x - (12))$

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

$y - 5 = - \frac{1}{2} \cdot (x - 12)$

Step 4

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

Solve for $y$ .

Simplify $- \frac{1}{2} \cdot (x - 12)$ .

Rewrite.

$y - 5 = 0 + 0 - \frac{1}{2} \cdot (x - 12)$

Simplify by adding zeros.

$y - 5 = - \frac{1}{2} \cdot (x - 12)$

Apply the distributive property.

$y - 5 = - \frac{1}{2} x - \frac{1}{2} \cdot - 12$

Combine $x$ and $\frac{1}{2}$ .

$y - 5 = - \frac{x}{2} - \frac{1}{2} \cdot - 12$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$y - 5 = - \frac{x}{2} + \frac{- 1}{2} \cdot - 12$

Factor $2$ out of $- 12$ .

$y - 5 = - \frac{x}{2} + \frac{- 1}{2} \cdot (2 (- 6))$

Cancel the common factor.

$y - 5 = - \frac{x}{2} + \frac{- 1}{2} \cdot (2 \cdot - 6)$

Rewrite the expression.

$y - 5 = - \frac{x}{2} - 1 \cdot - 6$

Multiply $- 1$ by $- 6$ .

$y - 5 = - \frac{x}{2} + 6$

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

Add $5$ to both sides of the equation.

$y = - \frac{x}{2} + 6 + 5$

Add $6$ and $5$ .

$y = - \frac{x}{2} + 11$

Reorder terms.

$y = - (\frac{1}{2} x) + 11$

Remove parentheses.

$y = - \frac{1}{2} x + 11$

Step 5

Find the Perpendicular Line y=2x+1 at (12,5)

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