How do you write an equation for the line in slope intercept form that is perpendicular to the given line and that passes through the given point 3x – 12y = –9; (1, –10)?

Jun 12, 2016

$y = - 4 x - 6$.

Explanation:

The slope of the given line is (using its standard eqn. $a x + b y + c = 0$)
$- \frac{a}{b} = - \left(\frac{3}{-} 12\right) = \frac{1}{4.}$

Hence, the slope of reqd. perpendicular line $= - \frac{1}{\frac{1}{4}} = - 4.$

The rqd. line is thro. $\left(1 , - 10\right)$.

Therefore, using slope-point form of reqd. line, is, $y - \left(- 10\right) = - 4 \left(x - 1\right) .$.
$\therefore y + 10 = - 4 x + 4$, or, $y = - 4 x - 6.$ (slope-intercept form)

Otherwise

A sort-cut method to find the eqn. of a line perpendicular to a given line $a x + b y + c = 0$ & passing thro. $\left({x}_{1} , {y}_{1}\right)$ is $b \left(x - {x}_{1}\right) - a \left(y - {y}_{1}\right) = 0.$

In our case, it is given by $- 12 \left(x - 1\right) - 3 \left\{y - \left(- 10\right)\right\} = 0.$
$\therefore y + 10 = - 4 \left(x - 1\right) = - 4 x + 4$, or, $y = - 4 x - 6$.