Determine the point on the line 5x-4y-2=0 where the normal that passes through the point (7,-2) intersects the line.
Attempt at a solution
The normal to y= mx+ b has slope -1/m.
5x- 4y- 2= 0 is the same as 4y= 5x- 2 or y= (5/4)x- 1/2. Its slope is 5/4 so the slope of any normal to it is -4/5.
What is the equation of the line through (7, -2) with slope -4/5?
r = (7,-2) + t (-4,5)
But where do I go from here? Can anyone point me in the right direction?
Sorry if I'm being a little frustrating, I had pneumonia and missed nearly all the lessons, and I'm really struggling to catch back up.