Question

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

Answer 1

`y=-16/11x-29/11`.

Explanation:

Any straight line graph as equation `y=mx+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=11/16x` has gradient `m=11/16`.
So another line perpendicular to this one will have gradient `-16/11`.
Now substituting the point `(x,y)=(3,-7)` into its general equation `y=mx+c` yields

`-7=(-16/11)(3)+c`,

from which we obtain the value for `c=-29/11`.

Thus the required line has equation `y=-16/11x-29/11`.