Find Any Equation Perpendicular to the Line y-3=4(x-10)

$y - 3 = 4 (x - 10)$

Step 1

Choose a point that the perpendicular line will pass through.

$(0, 0)$

Step 2

Solve $y - 3 = 4 (x - 10)$ .

Simplify $4 (x - 10)$ .

Rewrite.

$y - 3 = 0 + 0 + 4 (x - 10)$

Simplify by adding zeros.

$y - 3 = 4 (x - 10)$

Apply the distributive property.

$y - 3 = 4 x + 4 \cdot - 10$

Multiply $4$ by $- 10$ .

$y - 3 = 4 x - 40$

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

Add $3$ to both sides of the equation.

$y = 4 x - 40 + 3$

Add $- 40$ and $3$ .

$y = 4 x - 37$

Step 3

Use the slope-intercept form to find the slope.

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

$y = m x + b$

Using the slope-intercept form, the slope is $4$ .

$m = 4$

Step 4

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

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

Step 5

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

Use the slope $- \frac{1}{4}$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (0) = - \frac{1}{4} \cdot (x - (0))$

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

$y + 0 = - \frac{1}{4} \cdot (x + 0)$

Step 6

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

Solve for $y$ .

Add $y$ and $0$ .

$y = - \frac{1}{4} \cdot (x + 0)$

Simplify $- \frac{1}{4} \cdot (x + 0)$ .

Add $x$ and $0$ .

$y = - \frac{1}{4} \cdot x$

Combine $x$ and $\frac{1}{4}$ .

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

Reorder terms.

$y = - (\frac{1}{4} x)$

Remove parentheses.

$y = - \frac{1}{4} x$

Step 7

Find Any Equation Perpendicular to the Line y-3=4(x-10)

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