Question

Can someone please do this problem. Find k so that the line through (k-1,k+2) and (4,-1) is perpendicular to the line through (-3,2) and (2,5)

Answer 1

Can anyone solve this problem it is part of my math project

Answer 2

First find the gradient of the known line, which is (y2-y1)/(x2-x1) which comes out to 3/5. So we now know the gradient of the line we want is -5/3 because in order for the gradients to be perpendicular they must times together to = -1.

We know now a gradient and a point on the line so we can work out its equation \[y-y1 = m(x-x1)\] which comes out to y+1 = -5/3(x-4) which --> 3y+3 = -5(x-4) --> 3y = -5x + 20 -3 --> y+5x = 17

k+2 + 5(k-1) = 17
k+2 + 5k -5 = 17
6k = 17 + 3
6k = 20
k = 6/20
k = 3/10

(Think I went wrong somewhere but can't work out where, might be right but seems an odd number).

Answer 3

Where did you get
3y+3=-5(x-4) from

Answer 4

I multipled through out by 3 to get rid of the fraction?

Answer 5

Thanks this project is worth a test grade

Answer 6

Oh dear, I really hope that is all correct for ya and GL.