Find the Perpendicular Line What is an equation of the line that passes through the point (2,-3) and is perpendicular to the line 2x+5y=10 ?

What is an equation of the line that passes through the point $(2, - 3)$ and is perpendicular to the line $2 x + 5 y = 10$ ?

Step 1

Solve $2 x + 5 y = 10$ .

Subtract $2 x$ from both sides of the equation.

$5 y = 10 - 2 x$

Divide each term in $5 y = 10 - 2 x$ by $5$ and simplify.

Divide each term in $5 y = 10 - 2 x$ by $5$ .

$\frac{5 y}{5} = \frac{10}{5} + \frac{- 2 x}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 y}{5} = \frac{10}{5} + \frac{- 2 x}{5}$

Divide $y$ by $1$ .

$y = \frac{10}{5} + \frac{- 2 x}{5}$

Simplify the right side.

Simplify each term.

Divide $10$ by $5$ .

$y = 2 + \frac{- 2 x}{5}$

Move the negative in front of the fraction.

$y = 2 - \frac{2 x}{5}$

Step 2

Find the slope when $y = 2 - \frac{2 x}{5}$ .

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 $2$ and $- \frac{2 x}{5}$ .

$y = - \frac{2 x}{5} + 2$

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

Reorder terms.

$y = - (\frac{2}{5} x) + 2$

Remove parentheses.

$y = - \frac{2}{5} x + 2$

Using the slope-intercept form, the slope is $- \frac{2}{5}$ .

$m = - \frac{2}{5}$

Step 3

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

$m_{perpendicular} = - \frac{1}{- \frac{2}{5}}$

Step 4

Simplify $- \frac{1}{- \frac{2}{5}}$ to find the slope of the perpendicular line.

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

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

$m_{perpendicular} = - \frac{- 1 \cdot - 1}{- \frac{2}{5}}$

Move the negative in front of the fraction.

$m_{perpendicular} = \frac{1}{\frac{2}{5}}$

Multiply the numerator by the reciprocal of the denominator.

$m_{perpendicular} = 1 (\frac{5}{2})$

Multiply $\frac{5}{2}$ by $1$ .

$m_{perpendicular} = \frac{5}{2}$

Multiply $- - \frac{5}{2}$ .

Multiply $- 1$ by $- 1$ .

$m_{perpendicular} = 1 (\frac{5}{2})$

Multiply $\frac{5}{2}$ by $1$ .

$m_{perpendicular} = \frac{5}{2}$

Step 5

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

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

$y - (- 3) = \frac{5}{2} \cdot (x - (2))$

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

$y + 3 = \frac{5}{2} \cdot (x - 2)$

Step 6

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

Solve for $y$ .

Simplify $\frac{5}{2} \cdot (x - 2)$ .

Rewrite.

$y + 3 = 0 + 0 + \frac{5}{2} \cdot (x - 2)$

Simplify by adding zeros.

$y + 3 = \frac{5}{2} \cdot (x - 2)$

Apply the distributive property.

$y + 3 = \frac{5}{2} x + \frac{5}{2} \cdot - 2$

Combine $\frac{5}{2}$ and $x$ .

$y + 3 = \frac{5 x}{2} + \frac{5}{2} \cdot - 2$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

$y + 3 = \frac{5 x}{2} + \frac{5}{2} \cdot (2 (- 1))$

Cancel the common factor.

$y + 3 = \frac{5 x}{2} + \frac{5}{2} \cdot (2 \cdot - 1)$

Rewrite the expression.

$y + 3 = \frac{5 x}{2} + 5 \cdot - 1$

Multiply $5$ by $- 1$ .

$y + 3 = \frac{5 x}{2} - 5$

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

Subtract $3$ from both sides of the equation.

$y = \frac{5 x}{2} - 5 - 3$

Subtract $3$ from $- 5$ .

$y = \frac{5 x}{2} - 8$

Reorder terms.

$y = \frac{5}{2} x - 8$

Step 7

Find the Perpendicular Line What is an equation of the line that passes through the point (2,-3) and is perpendicular to the line 2x+5y=10 ?

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