Find the Perpendicular Line What is an equation of the line that passes through the point (6,-4) and is perpendicular to the line 2x-y=5

What is an equation of the line that passes through the point $(6, - 4)$ and is perpendicular to the line $2 x - y = 5$

Write the problem as a mathematical expression.

$(6, - 4)$ , $2 x - y = 5$

Solve $2 x - y = 5$ .

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

$- y = 5 - 2 x$

Divide each term in $- y = 5 - 2 x$ by $- 1$ and simplify.

Divide each term in $- y = 5 - 2 x$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Divide $5$ by $- 1$ .

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

Move the negative one from the denominator of $\frac{- 2 x}{- 1}$ .

$y = - 5 - 1 \cdot (- 2 x)$

Rewrite $- 1 \cdot (- 2 x)$ as $- (- 2 x)$ .

$y = - 5 - (- 2 x)$

Multiply $- 2$ by $- 1$ .

$y = - 5 + 2 x$

Find the slope when $y = - 5 + 2 x$ .

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 $- 5$ and $2 x$ .

$y = 2 x - 5$

Using the slope-intercept form, the slope is $2$ .

$m = 2$

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

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

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

Use the slope $- \frac{1}{2}$ and a given point $(6, - 4)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (- 4) = - \frac{1}{2} \cdot (x - (6))$

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

$y + 4 = - \frac{1}{2} \cdot (x - 6)$

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

Solve for $y$ .

Simplify $- \frac{1}{2} \cdot (x - 6)$ .

Rewrite.

$y + 4 = 0 + 0 - \frac{1}{2} \cdot (x - 6)$

Simplify by adding zeros.

$y + 4 = - \frac{1}{2} \cdot (x - 6)$

Apply the distributive property.

$y + 4 = - \frac{1}{2} x - \frac{1}{2} \cdot - 6$

Combine $x$ and $\frac{1}{2}$ .

$y + 4 = - \frac{x}{2} - \frac{1}{2} \cdot - 6$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$y + 4 = - \frac{x}{2} + \frac{- 1}{2} \cdot - 6$

Factor $2$ out of $- 6$ .

$y + 4 = - \frac{x}{2} + \frac{- 1}{2} \cdot (2 (- 3))$

Cancel the common factor.

$y + 4 = - \frac{x}{2} + \frac{- 1}{2} \cdot (2 \cdot - 3)$

Rewrite the expression.

$y + 4 = - \frac{x}{2} - 1 \cdot - 3$

Multiply $- 1$ by $- 3$ .

$y + 4 = - \frac{x}{2} + 3$

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

Subtract $4$ from both sides of the equation.

$y = - \frac{x}{2} + 3 - 4$

Subtract $4$ from $3$ .

$y = - \frac{x}{2} - 1$

Reorder terms.

$y = - (\frac{1}{2} x) - 1$

Remove parentheses.

$y = - \frac{1}{2} x - 1$

Find the Perpendicular Line What is an equation of the line that passes through the point (6,-4) and is perpendicular to the line 2x-y=5

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