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