Find the Perpendicular Line (5,-3) that is perpendicular to the line 5x+7y=8

$(5, - 3)$ that is perpendicular to the line $5 x + 7 y = 8$

Step 1

Solve $5 x + 7 y = 8$ .

Subtract $5 x$ from both sides of the equation.

$7 y = 8 - 5 x$

Divide each term in $7 y = 8 - 5 x$ by $7$ and simplify.

Divide each term in $7 y = 8 - 5 x$ by $7$ .

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

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 2

Find the slope when $y = \frac{8}{7} - \frac{5 x}{7}$ .

Rewrite in slope-intercept form.

The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

Reorder $\frac{8}{7}$ and $- \frac{5 x}{7}$ .

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

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

Reorder terms.

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

Remove parentheses.

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

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.

Cancel the common factor of $1$ and $- 1$ .

Rewrite $1$ as $- 1 (- 1)$ .

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

Move the negative in front of the fraction.

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $\frac{7}{5}$ by $1$ .

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

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

Multiply $- 1$ by $- 1$ .

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

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.

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

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

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.

Solve for $y$ .

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

Rewrite.

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

Simplify by adding zeros.

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

Apply the distributive property.

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

Combine $\frac{7}{5}$ and $x$ .

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

Cancel the common factor of $5$ .

Factor $5$ out of $- 5$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $7$ by $- 1$ .

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

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

Subtract $3$ from both sides of the equation.

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

Subtract $3$ from $- 7$ .

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

Reorder terms.

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

Step 7

Find the Perpendicular Line (5,-3) that is perpendicular to the line 5x+7y=8

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