Find the Perpendicular Line Passing through (8,-3) and perpendicular to the line whose equation is y=1/2x+2

Passing through $(8, - 3)$ and perpendicular to the line whose equation is $y = \frac{1}{2} x + 2$

Find the slope when $y = \frac{1}{2} x + 2$ .

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.

$y = m x + b$

Simplify the right side.

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

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

Reorder terms.

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

Using the slope-intercept form, the slope is $\frac{1}{2}$ .

$m = \frac{1}{2}$

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

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

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

Multiply the numerator by the reciprocal of the denominator.

$m_{perpendicular} = - (1 \cdot 2)$

Multiply $2$ by $1$ .

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

Multiply $- 1$ by $2$ .

$m_{perpendicular} = - 2$

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

Use the slope $- 2$ and a given point $(8, - 3)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (- 3) = - 2 \cdot (x - (8))$

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

$y + 3 = - 2 \cdot (x - 8)$

Solve for $y$ .

Simplify $- 2 \cdot (x - 8)$ .

Rewrite.

$y + 3 = 0 + 0 - 2 \cdot (x - 8)$

Simplify by adding zeros.

$y + 3 = - 2 \cdot (x - 8)$

Apply the distributive property.

$y + 3 = - 2 x - 2 \cdot - 8$

Multiply $- 2$ by $- 8$ .

$y + 3 = - 2 x + 16$

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

Subtract $3$ from both sides of the equation.

$y = - 2 x + 16 - 3$

Subtract $3$ from $16$ .

$y = - 2 x + 13$

Find the Perpendicular Line Passing through (8,-3) and perpendicular to the line whose equation is y=1/2x+2

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