Find Any Equation Perpendicular to the Line 3x+y=10

$3 x + y = 10$

Step 1

Choose a point that the perpendicular line will pass through.

$(0, 0)$

Step 2

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

$y = 10 - 3 x$

Step 3

Find the slope when $y = 10 - 3 x$ .

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 $10$ and $- 3 x$ .

$y = - 3 x + 10$

Using the slope-intercept form, the slope is $- 3$ .

$m = - 3$

Step 4

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

$m_{perpendicular} = - \frac{1}{- 3}$

Step 5

Simplify $- \frac{1}{- 3}$ to find the slope of the perpendicular line.

Move the negative in front of the fraction.

$m_{perpendicular} = \frac{1}{3}$

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

Multiply $- 1$ by $- 1$ .

$m_{perpendicular} = 1 (\frac{1}{3})$

Multiply $\frac{1}{3}$ by $1$ .

$m_{perpendicular} = \frac{1}{3}$

Step 6

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

Use the slope $\frac{1}{3}$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

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

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

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

Step 7

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

Solve for $y$ .

Add $y$ and $0$ .

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

Simplify $\frac{1}{3} \cdot (x + 0)$ .

Add $x$ and $0$ .

$y = \frac{1}{3} \cdot x$

Combine $\frac{1}{3}$ and $x$ .

$y = \frac{x}{3}$

Reorder terms.

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

Step 8

Find Any Equation Perpendicular to the Line 3x+y=10

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