Find the Perpendicular Line through: (3,5) , perp. to y=-3/7x-3

through: $(3, 5)$ , perp. to $y = - \frac{3}{7} x - 3$

Find the slope when $y = - \frac{3}{7} x - 3$ .

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.

Simplify each term.

Combine $x$ and $\frac{3}{7}$ .

$y = - \frac{x \cdot 3}{7} - 3$

Move $3$ to the left of $x$ .

$y = - \frac{3 x}{7} - 3$

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

Reorder terms.

$y = - (\frac{3}{7} x) - 3$

Remove parentheses.

$y = - \frac{3}{7} x - 3$

Using the slope-intercept form, the slope is $- \frac{3}{7}$ .

$m = - \frac{3}{7}$

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

$m_{perpendicular} = - \frac{1}{- \frac{3}{7}}$

Simplify $- \frac{1}{- \frac{3}{7}}$ to find the slope of the perpendicular line.

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

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

$m_{perpendicular} = - \frac{- 1 \cdot - 1}{- \frac{3}{7}}$

Move the negative in front of the fraction.

$m_{perpendicular} = \frac{1}{\frac{3}{7}}$

Multiply the numerator by the reciprocal of the denominator.

$m_{perpendicular} = 1 (\frac{7}{3})$

Multiply $\frac{7}{3}$ by $1$ .

$m_{perpendicular} = \frac{7}{3}$

Multiply $- - \frac{7}{3}$ .

Multiply $- 1$ by $- 1$ .

$m_{perpendicular} = 1 (\frac{7}{3})$

Multiply $\frac{7}{3}$ by $1$ .

$m_{perpendicular} = \frac{7}{3}$

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

Use the slope $\frac{7}{3}$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (5) = \frac{7}{3} \cdot (x - (3))$

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

$y - 5 = \frac{7}{3} \cdot (x - 3)$

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

Solve for $y$ .

Simplify $\frac{7}{3} \cdot (x - 3)$ .

Rewrite.

$y - 5 = 0 + 0 + \frac{7}{3} \cdot (x - 3)$

Simplify by adding zeros.

$y - 5 = \frac{7}{3} \cdot (x - 3)$

Apply the distributive property.

$y - 5 = \frac{7}{3} x + \frac{7}{3} \cdot - 3$

Combine $\frac{7}{3}$ and $x$ .

$y - 5 = \frac{7 x}{3} + \frac{7}{3} \cdot - 3$

Cancel the common factor of $3$ .

Factor $3$ out of $- 3$ .

$y - 5 = \frac{7 x}{3} + \frac{7}{3} \cdot (3 (- 1))$

Cancel the common factor.

$y - 5 = \frac{7 x}{3} + \frac{7}{3} \cdot (3 \cdot - 1)$

Rewrite the expression.

$y - 5 = \frac{7 x}{3} + 7 \cdot - 1$

Multiply $7$ by $- 1$ .

$y - 5 = \frac{7 x}{3} - 7$

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

Add $5$ to both sides of the equation.

$y = \frac{7 x}{3} - 7 + 5$

Add $- 7$ and $5$ .

$y = \frac{7 x}{3} - 2$

Reorder terms.

$y = \frac{7}{3} x - 2$

Find the Perpendicular Line through: (3,5) , perp. to y=-3/7x-3

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