Pre-Algebra Examples

y = x - 9

$y = x - 9$ ,

(0, 0)

$(0, 0)$

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

$y = m x + b$ , where

m

$m$ is the slope and

b

$b$ is the y-intercept.

y = m x + b

$y = m x + b$

Step 1.2

Using the slope-intercept form, the slope is

1

$1$ .

m = 1

$m = 1$

m = 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}

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

Step 3

Simplify

- \frac{1}{1}

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

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Step 3.1

Cancel the common factor of

1

$1$ .

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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

$(0, 0)$ 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

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (0) = - 1 \cdot (x - (0))$

Step 4.2

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

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

Step 5

Solve for

$y$ .

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Step 5.1

Add

$y$ and

$0$ .

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

Step 5.2

Simplify

$- 1 \cdot (x + 0)$ .

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Step 5.2.1

Add

$x$ and

$0$ .

$y = - 1 \cdot x$

Step 5.2.2

Rewrite

$- 1 x$ as

$- x$ .

$y = - x$

Step 6

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