Algebra Examples

17 x + 2 y = 0

$17 x + 2 y = 0$

Step 1

Choose a point that the perpendicular line will pass through.

(0, 0)

$(0, 0)$

Step 2

Solve

17 x + 2 y = 0

$17 x + 2 y = 0$ .

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Step 2.1

Subtract

17 x

$17 x$ from both sides of the equation.

2 y = - 17 x

$2 y = - 17 x$

Step 2.2

Divide each term in

2 y = - 17 x

$2 y = - 17 x$ by

2

$2$ and simplify.

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Step 2.2.1

Divide each term in

2 y = - 17 x

$2 y = - 17 x$ by

2

$2$ .

\frac{2 y}{2} = \frac{- 17 x}{2}

$\frac{2 y}{2} = \frac{- 17 x}{2}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of

2

$2$ .

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Step 2.2.2.1.1

Cancel the common factor.

\frac{2 y}{2} = \frac{- 17 x}{2}

$\frac{2 y}{2} = \frac{- 17 x}{2}$

Step 2.2.2.1.2

Divide

y

$y$ by

1

$1$ .

y = \frac{- 17 x}{2}

$y = \frac{- 17 x}{2}$

y = \frac{- 17 x}{2}

$y = \frac{- 17 x}{2}$

y = \frac{- 17 x}{2}

$y = \frac{- 17 x}{2}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Move the negative in front of the fraction.

y = - \frac{17 x}{2}

$y = - \frac{17 x}{2}$

y = - \frac{17 x}{2}

$y = - \frac{17 x}{2}$

y = - \frac{17 x}{2}

$y = - \frac{17 x}{2}$

y = - \frac{17 x}{2}

$y = - \frac{17 x}{2}$

Step 3

Find the slope when

y = - \frac{17 x}{2}

$y = - \frac{17 x}{2}$ .

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Step 3.1

Rewrite in slope-intercept form.

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Step 3.1.1

The slope-intercept form is

y = m x + b

$y = m x + b$ , where

m

$m$ is the slope and

b

$b$ is the y-intercept.

y = m x + b

$y = m x + b$

Step 3.1.2

Write in

y = m x + b

$y = m x + b$ form.

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Step 3.1.2.1

Reorder terms.

y = - (\frac{17}{2} x)

$y = - (\frac{17}{2} x)$

Step 3.1.2.2

Remove parentheses.

y = - \frac{17}{2} x

$y = - \frac{17}{2} x$

y = - \frac{17}{2} x

$y = - \frac{17}{2} x$

y = - \frac{17}{2} x

$y = - \frac{17}{2} x$

Step 3.2

Using the slope-intercept form, the slope is

- \frac{17}{2}

$- \frac{17}{2}$ .

m = - \frac{17}{2}

$m = - \frac{17}{2}$

m = - \frac{17}{2}

$m = - \frac{17}{2}$

Step 4

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

m_{perpendicular} = - \frac{1}{- \frac{17}{2}}

$m_{perpendicular} = - \frac{1}{- \frac{17}{2}}$

Step 5

Simplify

- \frac{1}{- \frac{17}{2}}

$- \frac{1}{- \frac{17}{2}}$ to find the slope of the perpendicular line.

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Step 5.1

Cancel the common factor of

1

$1$ and

- 1

$- 1$ .

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Step 5.1.1

Rewrite

1

$1$ as

- 1 (- 1)

$- 1 (- 1)$ .

m_{perpendicular} = - \frac{- 1 \cdot - 1}{- \frac{17}{2}}

$m_{perpendicular} = - \frac{- 1 \cdot - 1}{- \frac{17}{2}}$

Step 5.1.2

Move the negative in front of the fraction.

m_{perpendicular} = \frac{1}{\frac{17}{2}}

$m_{perpendicular} = \frac{1}{\frac{17}{2}}$

m_{perpendicular} = \frac{1}{\frac{17}{2}}

$m_{perpendicular} = \frac{1}{\frac{17}{2}}$

Step 5.2

Multiply the numerator by the reciprocal of the denominator.

m_{perpendicular} = 1 (\frac{2}{17})

$m_{perpendicular} = 1 (\frac{2}{17})$

Step 5.3

Multiply

\frac{2}{17}

$\frac{2}{17}$ by

1

$1$ .

m_{perpendicular} = \frac{2}{17}

$m_{perpendicular} = \frac{2}{17}$

Step 5.4

Multiply

- - \frac{2}{17}

$- - \frac{2}{17}$ .

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Step 5.4.1

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

m_{perpendicular} = 1 (\frac{2}{17})

$m_{perpendicular} = 1 (\frac{2}{17})$

Step 5.4.2

Multiply

\frac{2}{17}

$\frac{2}{17}$ by

1

$1$ .

m_{perpendicular} = \frac{2}{17}

$m_{perpendicular} = \frac{2}{17}$

m_{perpendicular} = \frac{2}{17}

$m_{perpendicular} = \frac{2}{17}$

m_{perpendicular} = \frac{2}{17}

$m_{perpendicular} = \frac{2}{17}$

Step 6

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

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Step 6.1

Use the slope

\frac{2}{17}

$\frac{2}{17}$ and a given point

(0, 0)

$(0, 0)$ to substitute for

x_{1}

$x_{1}$ and

y_{1}

$y_{1}$ in the point-slope form

y - y_{1} = m (x - x_{1})

$y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

y - (0) = \frac{2}{17} \cdot (x - (0))

$y - (0) = \frac{2}{17} \cdot (x - (0))$

Step 6.2

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

y + 0 = \frac{2}{17} \cdot (x + 0)

$y + 0 = \frac{2}{17} \cdot (x + 0)$

y + 0 = \frac{2}{17} \cdot (x + 0)

$y + 0 = \frac{2}{17} \cdot (x + 0)$

Step 7

Write in

y = m x + b

$y = m x + b$ form.

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Step 7.1

Solve for

y

$y$ .

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Step 7.1.1

Add

y

$y$ and

0

$0$ .

y = \frac{2}{17} \cdot (x + 0)

$y = \frac{2}{17} \cdot (x + 0)$

Step 7.1.2

Simplify

\frac{2}{17} \cdot (x + 0)

$\frac{2}{17} \cdot (x + 0)$ .

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Step 7.1.2.1

Add

x

$x$ and

0

$0$ .

y = \frac{2}{17} \cdot x

$y = \frac{2}{17} \cdot x$

Step 7.1.2.2

Combine

\frac{2}{17}

$\frac{2}{17}$ and

x

$x$ .

y = \frac{2 x}{17}

$y = \frac{2 x}{17}$

y = \frac{2 x}{17}

$y = \frac{2 x}{17}$

y = \frac{2 x}{17}

$y = \frac{2 x}{17}$

Step 7.2

Reorder terms.

y = \frac{2}{17} x

$y = \frac{2}{17} x$

y = \frac{2}{17} x

$y = \frac{2}{17} x$

Step 8

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