# What is the equation of the line that passes through (-2,1)  and is perpendicular to the line that passes through the following points: #(-16,4),(6,12)?

Aug 1, 2016

Let's first find the equation of the line that it's perpendicular to. We need to find the slope for this:

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

$m = \frac{12 - 4}{6 - \left(- 16\right)}$

$m = \frac{8}{22}$

$m = \frac{4}{11}$

Now, by point-slope form:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 12 = \frac{4}{11} \left(x - 6\right)$

$y - 12 = \frac{4}{11} x - \frac{24}{11}$

$y = \frac{4}{11} x - \frac{24}{11} + 12$

$y = \frac{4}{11} x + \frac{108}{11}$

The slope of a line perpendicular to another always has a slope that is the negative reciprocal of the other line.

Hence, ${m}_{\text{perpendicular}} = - \frac{11}{4}$

Again, by point-slope form:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 1 = - \frac{11}{4} \left(x - \left(- 2\right)\right)$

$y - 1 = - \frac{11}{4} x - \frac{11}{2}$

$y = - \frac{11}{4} x - \frac{11}{2} + 1$

$y = - \frac{11}{4} x - \frac{9}{2}$

$\therefore$The equation of the line is $y = - \frac{11}{4} x - \frac{9}{2}$.

Hopefully this helps!