Find the Perpendicular Line Through (1/2,-2/3) ; perpendicular to the line 5x-10y=1

Through $(\frac{1}{2}, - \frac{2}{3})$ ; perpendicular to the line $5 x - 10 y = 1$

Solve $5 x - 10 y = 1$ .

Subtract $5 x$ from both sides of the equation.

$- 10 y = 1 - 5 x$

Divide each term in $- 10 y = 1 - 5 x$ by $- 10$ and simplify.

Divide each term in $- 10 y = 1 - 5 x$ by $- 10$ .

$\frac{- 10 y}{- 10} = \frac{1}{- 10} + \frac{- 5 x}{- 10}$

Simplify the left side.

Cancel the common factor of $- 10$ .

Cancel the common factor.

$\frac{- 10 y}{- 10} = \frac{1}{- 10} + \frac{- 5 x}{- 10}$

Divide $y$ by $1$ .

$y = \frac{1}{- 10} + \frac{- 5 x}{- 10}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y = - \frac{1}{10} + \frac{- 5 x}{- 10}$

Cancel the common factor of $- 5$ and $- 10$ .

Factor $- 5$ out of $- 5 x$ .

$y = - \frac{1}{10} + \frac{- 5 (x)}{- 10}$

Cancel the common factors.

Factor $- 5$ out of $- 10$ .

$y = - \frac{1}{10} + \frac{- 5 (x)}{- 5 (2)}$

Cancel the common factor.

$y = - \frac{1}{10} + \frac{- 5 x}{- 5 \cdot 2}$

Rewrite the expression.

$y = - \frac{1}{10} + \frac{x}{2}$

Find the slope when $y = - \frac{1}{10} + \frac{x}{2}$ .

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.

$y = m x + b$

Reorder $- \frac{1}{10}$ and $\frac{x}{2}$ .

$y = \frac{x}{2} - \frac{1}{10}$

Reorder terms.

$y = \frac{1}{2} x - \frac{1}{10}$

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 $(\frac{1}{2}, - \frac{2}{3})$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (- \frac{2}{3}) = - 2 \cdot (x - (\frac{1}{2}))$

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

$y + \frac{2}{3} = - 2 \cdot (x - \frac{1}{2})$

Solve for $y$ .

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

Rewrite.

$y + \frac{2}{3} = 0 + 0 - 2 \cdot (x - \frac{1}{2})$

Simplify by adding zeros.

$y + \frac{2}{3} = - 2 \cdot (x - \frac{1}{2})$

Apply the distributive property.

$y + \frac{2}{3} = - 2 x - 2 (- \frac{1}{2})$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$y + \frac{2}{3} = - 2 x - 2 (\frac{- 1}{2})$

Factor $2$ out of $- 2$ .

$y + \frac{2}{3} = - 2 x + 2 (- 1) \frac{- 1}{2}$

Cancel the common factor.

$y + \frac{2}{3} = - 2 x + 2 \cdot - 1 \frac{- 1}{2}$

Rewrite the expression.

$y + \frac{2}{3} = - 2 x - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

$y + \frac{2}{3} = - 2 x + 1$

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

Subtract $\frac{2}{3}$ from both sides of the equation.

$y = - 2 x + 1 - \frac{2}{3}$

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Subtract $2$ from $3$ .

$y = - 2 x + \frac{1}{3}$

Find the Perpendicular Line Through (1/2,-2/3) ; perpendicular to the line 5x-10y=1

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