Question

What is the equation of the line perpendicular to Y-3x+1 that passes through the point (12,-6)?

Answer 1

Answer:

$3y + x = -6$

Step-by-step explanation:

Given

y = 3x + 1

Required

Equation of line that passes through (12,-6) and is perpendicular to y = 3x + 1

First, the slope of the line has to be calculated;

Th slope of a line is the coefficient of x in its linear equation;

This implies that the slope of y = 3x + 1 is 3

Having calculated the slope of the first line;

The relationship between both lines are perpendicularity; this implies that $m_1 * m_2 = -1$

Where m_1 = 3 and m_2 is the slope of the secodnd line

$m_1 * m_2 = -1$ becomes

$3 * m_2 = -1$

Divide both sides by 3

$\frac{3 *m_2}{m_2} = \frac{-1}{3}$

$m_2 = \frac{-1}{3}$

The equation of the line can be calculated using the folloing formula

$m = \frac{y - y_1}{x - x_1}$

Where $(x_1, y_1) = (12,-6)$ and $m_2 = \frac{-1}{3}$

The equation becomes

$\frac{-1}{3} = \frac{y -- 6}{x - 12}$

$\frac{-1}{3} = \frac{y + 6}{x - 12}$

Cross multiply

$-(x - 12)= 3(y + 6)$

$-x + 12=3y + 18$

Collect like terms

$12 - 18 = 3y +x$

$-6 = 3y + x$

$3y + x = -6$