Find the Perpendicular Line Passing through (3,-5) and perpendicular to the line whose equation is y=1/5x+2

Passing through $(3, - 5)$ and perpendicular to the line whose equation is $y = \frac{1}{5} x + 2$

Find the slope when $y = \frac{1}{5} x + 2$ .

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 $\frac{1}{5}$ and $x$ .

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

Reorder terms.

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

Using the slope-intercept form, the slope is $\frac{1}{5}$ .

$m = \frac{1}{5}$

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

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

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

Multiply the numerator by the reciprocal of the denominator.

$m_{perpendicular} = - (1 \cdot 5)$

Multiply $5$ by $1$ .

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

Multiply $- 1$ by $5$ .

$m_{perpendicular} = - 5$

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

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

$y - (- 5) = - 5 \cdot (x - (3))$

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

$y + 5 = - 5 \cdot (x - 3)$

Solve for $y$ .

Simplify $- 5 \cdot (x - 3)$ .

Rewrite.

$y + 5 = 0 + 0 - 5 \cdot (x - 3)$

Simplify by adding zeros.

$y + 5 = - 5 \cdot (x - 3)$

Apply the distributive property.

$y + 5 = - 5 x - 5 \cdot - 3$

Multiply $- 5$ by $- 3$ .

$y + 5 = - 5 x + 15$

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

Subtract $5$ from both sides of the equation.

$y = - 5 x + 15 - 5$

Subtract $5$ from $15$ .

$y = - 5 x + 10$

Find the Perpendicular Line Passing through (3,-5) and perpendicular to the line whose equation is y=1/5x+2

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