Precalculus Examples

$17 x + 2 y = 0$

Choose a point that the parallel line will pass through.

$(1, 0)$

Rewrite in slope-intercept form.

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The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

Subtract $17 x$ from both sides of the equation.

$2 y = - 17 x$

Divide each term by $2$ and simplify.

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Divide each term in $2 y = - 17 x$ by $2$ .

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

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

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

Divide $y$ by $1$ .

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

Move the negative in front of the fraction.

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

Rewrite in slope-intercept form.

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

Using the slope-intercept form, the slope is $- \frac{17}{2}$ .

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

To find an equation that is parallel to $17 x + 2 y = 0$ , the slopes must be equal. Using the slope of the equation, find the parallel line using the point-slope formula.

$(1, 0)$

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

Using the point-slope form $y - y_{1} = m (x - x_{1})$ , plug in $m = - \frac{17}{2}$ , $x_{1} = 1$ , and $y_{1} = 0$ .

$y - (0) = (- \frac{17}{2}) (x - (1))$

Solve for $y$ .

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Subtract $0$ from $y$ .

$y = (- \frac{17}{2}) (x - (1))$

Simplify $(- \frac{17}{2}) (x - (1))$ .

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Multiply $- 1$ by $1$ .

$y = - \frac{17}{2} \cdot (x - 1)$

Apply the distributive property.

$y = - \frac{17}{2} x - \frac{17}{2} \cdot - 1$

Combine $x$ and $\frac{17}{2}$ .

$y = - \frac{x \cdot 17}{2} - \frac{17}{2} \cdot - 1$

Multiply $- \frac{17}{2} \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$y = - \frac{x \cdot 17}{2} + 1 (\frac{17}{2})$

Multiply $\frac{17}{2}$ by $1$ .

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

Move $17$ to the left of $x$ .

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

Reorder terms.

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

Remove parentheses.

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

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