What is the equation of the line perpendicular to y=11/16x  that passes through  (3,-7) ?

Dec 29, 2015

$y = - \frac{16}{11} x - \frac{29}{11}$.

Explanation:

Any straight line graph as equation $y = m x + c$ where $m$ is the slope or gradient and $c$ is the y-intercept.

Furthermore, perpendicular lines have gradients whose products is $- 1$.

So in this case, $y = \frac{11}{16} x$ has gradient $m = \frac{11}{16}$.
So another line perpendicular to this one will have gradient $- \frac{16}{11}$.
Now substituting the point $\left(x , y\right) = \left(3 , - 7\right)$ into its general equation $y = m x + c$ yields

$- 7 = \left(- \frac{16}{11}\right) \left(3\right) + c$,

from which we obtain the value for $c = - \frac{29}{11}$.

Thus the required line has equation $y = - \frac{16}{11} x - \frac{29}{11}$.