Precalculus Examples Algebra Concepts and Expressions

Find the Perpendicular Line

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

Step 1

Use the slope-intercept form to find the slope.

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Step 1.1

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

$y = m x + b$

Step 1.2

Using the slope-intercept form, the slope is $1$ .

$m = 1$

Step 2

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

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

Step 3

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

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Step 3.1

Cancel the common factor of $1$ .

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Step 3.1.1

Cancel the common factor.

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

Step 3.1.2

Rewrite the expression.

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

Step 3.2

Multiply $- 1$ by $1$ .

$m_{perpendicular} = - 1$

Step 4

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

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Step 4.1

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

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

Step 4.2

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

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

Step 5

Solve for $y$ .

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Step 5.1

Simplify $- 1 \cdot (x - 2)$ .

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Step 5.1.1

Rewrite.

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

Step 5.1.2

Simplify by adding zeros.

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

Step 5.1.3

Apply the distributive property.

$y - 6 = - 1 x - 1 \cdot - 2$

Step 5.1.4

Simplify the expression.

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Step 5.1.4.1

Rewrite $- 1 x$ as $- x$ .

$y - 6 = - x - 1 \cdot - 2$

Step 5.1.4.2

Multiply $- 1$ by $- 2$ .

$y - 6 = - x + 2$

Step 5.2

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

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Step 5.2.1

Add $6$ to both sides of the equation.

$y = - x + 2 + 6$

Step 5.2.2

Add $2$ and $6$ .

$y = - x + 8$

Step 6

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Find the Perpendicular Line

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