Algebra Examples

(5, - 3)

$(5, - 3)$ that is perpendicular to the line

$5 x + 7 y = 8$

Step 1

Solve

$5 x + 7 y = 8$ .

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Step 1.1

Subtract

$5 x$ from both sides of the equation.

$7 y = 8 - 5 x$

Step 1.2

Divide each term in

$7 y = 8 - 5 x$ by

$7$ and simplify.

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Step 1.2.1

Divide each term in

$7 y = 8 - 5 x$ by

$7$ .

$\frac{7 y}{7} = \frac{8}{7} + \frac{- 5 x}{7}$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Cancel the common factor of

$7$ .

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Step 1.2.2.1.1

Cancel the common factor.

$\frac{7 y}{7} = \frac{8}{7} + \frac{- 5 x}{7}$

Step 1.2.2.1.2

Divide

$y$ by

$1$ .

$y = \frac{8}{7} + \frac{- 5 x}{7}$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Move the negative in front of the fraction.

$y = \frac{8}{7} - \frac{5 x}{7}$

Step 2

Find the slope when

$y = \frac{8}{7} - \frac{5 x}{7}$ .

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Step 2.1

Rewrite in slope-intercept form.

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Step 2.1.1

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 2.1.2

Reorder

$\frac{8}{7}$ and

$- \frac{5 x}{7}$ .

$y = - \frac{5 x}{7} + \frac{8}{7}$

Step 2.1.3

Write in

$y = m x + b$ form.

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Step 2.1.3.1

Reorder terms.

$y = - (\frac{5}{7} x) + \frac{8}{7}$

Step 2.1.3.2

Remove parentheses.

$y = - \frac{5}{7} x + \frac{8}{7}$

Step 2.2

Using the slope-intercept form, the slope is

$- \frac{5}{7}$ .

$m = - \frac{5}{7}$

Step 3

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

$m_{perpendicular} = - \frac{1}{- \frac{5}{7}}$

Step 4

Simplify

$- \frac{1}{- \frac{5}{7}}$ to find the slope of the perpendicular line.

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Step 4.1

Cancel the common factor of

$1$ and

$- 1$ .

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Step 4.1.1

Rewrite

$1$ as

$- 1 (- 1)$ .

$m_{perpendicular} = - \frac{- 1 \cdot - 1}{- \frac{5}{7}}$

Step 4.1.2

Move the negative in front of the fraction.

$m_{perpendicular} = \frac{1}{\frac{5}{7}}$

Step 4.2

Multiply the numerator by the reciprocal of the denominator.

$m_{perpendicular} = 1 (\frac{7}{5})$

Step 4.3

Multiply

$\frac{7}{5}$ by

$1$ .

$m_{perpendicular} = \frac{7}{5}$

Step 4.4

Multiply

$- - \frac{7}{5}$ .

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Step 4.4.1

Multiply

$- 1$ by

$- 1$ .

$m_{perpendicular} = 1 (\frac{7}{5})$

Step 4.4.2

Multiply

$\frac{7}{5}$ by

$1$ .

$m_{perpendicular} = \frac{7}{5}$

Step 5

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

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Step 5.1

Use the slope

$\frac{7}{5}$ and a given point

$(5, - 3)$ 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

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

$y - (- 3) = \frac{7}{5} \cdot (x - (5))$

Step 5.2

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

$y + 3 = \frac{7}{5} \cdot (x - 5)$

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

$\frac{7}{5} \cdot (x - 5)$ .

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Step 6.1.1.1

Rewrite.

$y + 3 = 0 + 0 + \frac{7}{5} \cdot (x - 5)$

Step 6.1.1.2

Simplify by adding zeros.

$y + 3 = \frac{7}{5} \cdot (x - 5)$

Step 6.1.1.3

Apply the distributive property.

$y + 3 = \frac{7}{5} x + \frac{7}{5} \cdot - 5$

Step 6.1.1.4

Combine

$\frac{7}{5}$ and

$x$ .

$y + 3 = \frac{7 x}{5} + \frac{7}{5} \cdot - 5$

Step 6.1.1.5

Cancel the common factor of

$5$ .

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Step 6.1.1.5.1

Factor

$5$ out of

$- 5$ .

$y + 3 = \frac{7 x}{5} + \frac{7}{5} \cdot (5 (- 1))$

Step 6.1.1.5.2

Cancel the common factor.

$y + 3 = \frac{7 x}{5} + \frac{7}{5} \cdot (5 \cdot - 1)$

Step 6.1.1.5.3

Rewrite the expression.

$y + 3 = \frac{7 x}{5} + 7 \cdot - 1$

Step 6.1.1.6

Multiply

$7$ by

$- 1$ .

$y + 3 = \frac{7 x}{5} - 7$

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

Subtract

$3$ from both sides of the equation.

$y = \frac{7 x}{5} - 7 - 3$

Step 6.1.2.2

Subtract

$3$ from

$- 7$ .

$y = \frac{7 x}{5} - 10$

Step 6.2

Reorder terms.

$y = \frac{7}{5} x - 10$

Step 7