Answer:
[tex]3y + x = -6[/tex]
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 [tex]m_1 * m_2 = -1[/tex]
Where m_1 = 3 and m_2 is the slope of the secodnd line
[tex]m_1 * m_2 = -1[/tex] becomes
[tex]3 * m_2 = -1[/tex]
Divide both sides by 3
[tex]\frac{3 *m_2}{m_2} = \frac{-1}{3}[/tex]
[tex]m_2 = \frac{-1}{3}[/tex]
The equation of the line can be calculated using the folloing formula
[tex]m = \frac{y - y_1}{x - x_1}[/tex]
Where [tex](x_1, y_1) = (12,-6)[/tex] and [tex]m_2 = \frac{-1}{3}[/tex]
The equation becomes
[tex]\frac{-1}{3} = \frac{y -- 6}{x - 12}[/tex]
[tex]\frac{-1}{3} = \frac{y + 6}{x - 12}[/tex]
Cross multiply
[tex]-(x - 12)= 3(y + 6)[/tex]
[tex]-x + 12=3y + 18[/tex]
Collect like terms
[tex]12 - 18 = 3y +x[/tex]
[tex]-6 = 3y + x[/tex]
[tex]3y + x = -6[/tex]
