Algebra Examples

$y - 5 = - 3 (x - 1)$ ; $(1, 5)$

Step 1

Solve $y - 5 = - 3 (x - 1)$ .

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Step 1.1

Simplify $- 3 (x - 1)$ .

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Step 1.1.1

Rewrite.

$y - 5 = 0 + 0 - 3 (x - 1)$

Step 1.1.2

Simplify by adding zeros.

$y - 5 = - 3 (x - 1)$

Step 1.1.3

Apply the distributive property.

$y - 5 = - 3 x - 3 \cdot - 1$

Step 1.1.4

Multiply $- 3$ by $- 1$ .

$y - 5 = - 3 x + 3$

Step 1.2

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

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Step 1.2.1

Add $5$ to both sides of the equation.

$y = - 3 x + 3 + 5$

Step 1.2.2

Add $3$ and $5$ .

$y = - 3 x + 8$

Step 2

Use the slope-intercept form to find the slope.

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Step 2.1

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 2.2

Using the slope-intercept form, the slope is $- 3$ .

$m = - 3$

Step 3

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

$m_{perpendicular} = - \frac{1}{- 3}$

Step 4

Simplify $- \frac{1}{- 3}$ to find the slope of the perpendicular line.

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Step 4.1

Move the negative in front of the fraction.

$m_{perpendicular} = \frac{1}{3}$

Step 4.2

Multiply $- - \frac{1}{3}$ .

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Step 4.2.1

Multiply $- 1$ by $- 1$ .

$m_{perpendicular} = 1 (\frac{1}{3})$

Step 4.2.2

Multiply $\frac{1}{3}$ by $1$ .

$m_{perpendicular} = \frac{1}{3}$

Step 5

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

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Step 5.1

Use the slope $\frac{1}{3}$ and a given point $(1, 5)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (5) = \frac{1}{3} \cdot (x - (1))$

Step 5.2

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

$y - 5 = \frac{1}{3} \cdot (x - 1)$

Step 6

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

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Step 6.1

Solve for $y$ .

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Step 6.1.1

Simplify $\frac{1}{3} \cdot (x - 1)$ .

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Step 6.1.1.1

Rewrite.

$y - 5 = 0 + 0 + \frac{1}{3} \cdot (x - 1)$

Step 6.1.1.2

Simplify by adding zeros.

$y - 5 = \frac{1}{3} \cdot (x - 1)$

Step 6.1.1.3

Apply the distributive property.

$y - 5 = \frac{1}{3} x + \frac{1}{3} \cdot - 1$

Step 6.1.1.4

Combine $\frac{1}{3}$ and $x$ .

$y - 5 = \frac{x}{3} + \frac{1}{3} \cdot - 1$

Step 6.1.1.5

Combine $\frac{1}{3}$ and $- 1$ .

$y - 5 = \frac{x}{3} + \frac{- 1}{3}$

Step 6.1.1.6

Move the negative in front of the fraction.

$y - 5 = \frac{x}{3} - \frac{1}{3}$

Step 6.1.2

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

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Step 6.1.2.1

Add $5$ to both sides of the equation.

$y = \frac{x}{3} - \frac{1}{3} + 5$

Step 6.1.2.2

To write $5$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = \frac{x}{3} - \frac{1}{3} + 5 \cdot \frac{3}{3}$

Step 6.1.2.3

Combine $5$ and $\frac{3}{3}$ .

$y = \frac{x}{3} - \frac{1}{3} + \frac{5 \cdot 3}{3}$

Step 6.1.2.4

Combine the numerators over the common denominator.

$y = \frac{x}{3} + \frac{- 1 + 5 \cdot 3}{3}$

Step 6.1.2.5

Simplify the numerator.

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Step 6.1.2.5.1

Multiply $5$ by $3$ .

$y = \frac{x}{3} + \frac{- 1 + 15}{3}$

Step 6.1.2.5.2

Add $- 1$ and $15$ .

$y = \frac{x}{3} + \frac{14}{3}$

Step 6.2

Reorder terms.

$y = \frac{1}{3} x + \frac{14}{3}$

Step 7