Find the Perpendicular Line (5,0),y+1=2(x-3)

$(5, 0), y + 1 = 2 (x - 3)$

Solve $y + 1 = 2 (x - 3)$ .

Simplify $2 (x - 3)$ .

Rewrite.

$y + 1 = 0 + 0 + 2 (x - 3)$

Simplify by adding zeros.

$y + 1 = 2 (x - 3)$

Apply the distributive property.

$y + 1 = 2 x + 2 \cdot - 3$

Multiply $2$ by $- 3$ .

$y + 1 = 2 x - 6$

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

Subtract $1$ from both sides of the equation.

$y = 2 x - 6 - 1$

Subtract $1$ from $- 6$ .

$y = 2 x - 7$

Use the slope-intercept form to find the slope.

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

$y = m x + b$

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 $(5, 0)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (0) = - \frac{1}{2} \cdot (x - (5))$

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

$y + 0 = - \frac{1}{2} \cdot (x - 5)$

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

Solve for $y$ .

Add $y$ and $0$ .

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

Simplify $- \frac{1}{2} \cdot (x - 5)$ .

Apply the distributive property.

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

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

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

Multiply $- \frac{1}{2} \cdot - 5$ .

Multiply $- 5$ by $- 1$ .

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

Combine $5$ and $\frac{1}{2}$ .

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

Reorder terms.

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

Remove parentheses.

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

Find the Perpendicular Line (5,0),y+1=2(x-3)

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