Find the Perpendicular Line line passing through (1,9) and perpendicular to 3x+7y=1

line passing through $(1, 9)$ and perpendicular to $3 x + 7 y = 1$

Solve $3 x + 7 y = 1$ .

Subtract $3 x$ from both sides of the equation.

$7 y = 1 - 3 x$

Divide each term in $7 y = 1 - 3 x$ by $7$ and simplify.

Divide each term in $7 y = 1 - 3 x$ by $7$ .

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

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Find the slope when $y = \frac{1}{7} - \frac{3 x}{7}$ .

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 $\frac{1}{7}$ and $- \frac{3 x}{7}$ .

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

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

Reorder terms.

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

Remove parentheses.

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

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 $(1, 9)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (9) = \frac{7}{3} \cdot (x - (1))$

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

$y - 9 = \frac{7}{3} \cdot (x - 1)$

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

Solve for $y$ .

Simplify $\frac{7}{3} \cdot (x - 1)$ .

Rewrite.

$y - 9 = 0 + 0 + \frac{7}{3} \cdot (x - 1)$

Simplify by adding zeros.

$y - 9 = \frac{7}{3} \cdot (x - 1)$

Apply the distributive property.

$y - 9 = \frac{7}{3} x + \frac{7}{3} \cdot - 1$

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

$y - 9 = \frac{7 x}{3} + \frac{7}{3} \cdot - 1$

Multiply $\frac{7}{3} \cdot - 1$ .

Combine $\frac{7}{3}$ and $- 1$ .

$y - 9 = \frac{7 x}{3} + \frac{7 \cdot - 1}{3}$

Multiply $7$ by $- 1$ .

$y - 9 = \frac{7 x}{3} + \frac{- 7}{3}$

Move the negative in front of the fraction.

$y - 9 = \frac{7 x}{3} - \frac{7}{3}$

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

Add $9$ to both sides of the equation.

$y = \frac{7 x}{3} - \frac{7}{3} + 9$

To write $9$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = \frac{7 x}{3} - \frac{7}{3} + 9 \cdot \frac{3}{3}$

Combine $9$ and $\frac{3}{3}$ .

$y = \frac{7 x}{3} - \frac{7}{3} + \frac{9 \cdot 3}{3}$

Combine the numerators over the common denominator.

$y = \frac{7 x}{3} + \frac{- 7 + 9 \cdot 3}{3}$

Simplify the numerator.

Multiply $9$ by $3$ .

$y = \frac{7 x}{3} + \frac{- 7 + 27}{3}$

Add $- 7$ and $27$ .

$y = \frac{7 x}{3} + \frac{20}{3}$

Reorder terms.

$y = \frac{7}{3} x + \frac{20}{3}$

Find the Perpendicular Line line passing through (1,9) and perpendicular to 3x+7y=1

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