Algebra Examples

Find the equation of a line parallel to

$y = 2 + 6 x$ that passes through the point

$(5, 0)$

Step 1

Find the slope when

$y = 2 + 6 x$ .

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Step 1.1

Rewrite in slope-intercept form.

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Step 1.1.1

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 1.1.2

Reorder

$2$ and

$6 x$ .

$y = 6 x + 2$

Step 1.2

Using the slope-intercept form, the slope is

$6$ .

$m = 6$

Step 2

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

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

Step 3

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

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Step 3.1

Use the slope

$- \frac{1}{6}$ and a given point

$(5, 0)$ 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

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

$y - (0) = - \frac{1}{6} \cdot (x - (5))$

Step 3.2

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

$y + 0 = - \frac{1}{6} \cdot (x - 5)$

Step 4

Write in

$y = m x + b$ form.

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Step 4.1

Solve for

$y$ .

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Step 4.1.1

Add

$y$ and

$0$ .

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

Step 4.1.2

Simplify

$- \frac{1}{6} \cdot (x - 5)$ .

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Step 4.1.2.1

Apply the distributive property.

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

Step 4.1.2.2

Combine

$x$ and

$\frac{1}{6}$ .

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

Step 4.1.2.3

Multiply

$- \frac{1}{6} \cdot - 5$ .

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Step 4.1.2.3.1

Multiply

$- 5$ by

$- 1$ .

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

Step 4.1.2.3.2

Combine

$5$ and

$\frac{1}{6}$ .

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

Step 4.2

Reorder terms.

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

Step 4.3

Remove parentheses.

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

Step 5