Find Any Equation Perpendicular to the Line y=4/7x-3

$y = \frac{4}{7} x - 3$

Step 1

Choose a point that the perpendicular line will pass through.

$(0, 0)$

Step 2

Find the slope when $y = \frac{4}{7} x - 3$ .

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.

Combine $\frac{4}{7}$ and $x$ .

$y = \frac{4 x}{7} - 3$

Reorder terms.

$y = \frac{4}{7} x - 3$

Using the slope-intercept form, the slope is $\frac{4}{7}$ .

$m = \frac{4}{7}$

Step 3

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

$m_{perpendicular} = - \frac{1}{\frac{4}{7}}$

Step 4

Simplify $- \frac{1}{\frac{4}{7}}$ to find the slope of the perpendicular line.

Multiply the numerator by the reciprocal of the denominator.

$m_{perpendicular} = - (1 (\frac{7}{4}))$

Multiply $\frac{7}{4}$ by $1$ .

$m_{perpendicular} = - \frac{7}{4}$

Step 5

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

Use the slope $- \frac{7}{4}$ and a given point $(0, 0)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (0) = - \frac{7}{4} \cdot (x - (0))$

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

$y + 0 = - \frac{7}{4} \cdot (x + 0)$

Step 6

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

Solve for $y$ .

Add $y$ and $0$ .

$y = - \frac{7}{4} \cdot (x + 0)$

Simplify $- \frac{7}{4} \cdot (x + 0)$ .

Add $x$ and $0$ .

$y = - \frac{7}{4} \cdot x$

Combine $x$ and $\frac{7}{4}$ .

$y = - \frac{x \cdot 7}{4}$

Move $7$ to the left of $x$ .

$y = - \frac{7 x}{4}$

Reorder terms.

$y = - (\frac{7}{4} x)$

Remove parentheses.

$y = - \frac{7}{4} x$

Step 7

Find Any Equation Perpendicular to the Line y=4/7x-3

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