What is the equation of the line that passes through (-4,1) and is perpendicular to the line that passes through the following points: (6,9),(-2,6) ?

Mar 7, 2016

$8 x + 3 y + 29 = 0$

Explanation:

Slope of the line that passes through points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by $\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$. Hence slope of line joining $\left(6 , 9\right)$ and (−2,6) is $\frac{6 - 9}{\left(- 2\right) - 6}$ or $- \frac{3}{-} 8 = \frac{3}{8}$.

As product of slope of two perpendicular lines is $- 1$, slope of line perpendicular to one joining $\left(6 , 9\right)$ and (−2,6) is $- \frac{1}{\frac{3}{8}}$ or $- \frac{8}{3}$.

Now using Point-slope form, the equation of line passing through $\left(- 4 , 1\right)$ and slope $- \frac{8}{3}$ will be

$\left(y - 1\right) = - \frac{8}{3} \times \left(x - \left(- 4\right)\right)$ or

$3 \left(y - 1\right) = \left(- 8\right) \times \left(x + 4\right)$ or

$3 y - 3 = - 8 x - 32$ or $8 x + 3 y + 29 = 0$