Question

Find the point at which the normal through the point (3,-4) to the line 10x+4y-101=0 intersects the line?

Answer 1

what is the slope of \(10x+4y-101=0 \) ?

Answer 2

Well, the normal vector is (10,4) the direction vector could them be (-4,10) or (4,-10), so the slope would be -5/4

Answer 3

then*

Answer 4

so you can find the equation of the line since you have a point and a slope
then find where the lines intersect

Answer 5

hmm, alright, i'll try.
thanks!

Answer 6

yw

Answer 7

Essentially by doing that, I'm saying the slope of both lines are the same, therefore they're parallel so they wouldn't intersect. But that's not true.

Answer 8

pretty much what you're being asked is
find the equation of the line that is perpendicular, "normal", to 10x+4y-101=0 and passes through (3, -4)

so you once you find that, you'd have 2 equations as satellite73 already pointed out
that is, a system of equations with two variables

recall that the solution to a system of equations is where their graph meet or intersect

the solution to the system of equations, is where the perpendicular line and the given line meet