Question

PLEASE PLEASE HELP. THIS IS URGENT

Find the value of K if the lines kx-3y=4 and 4x-(k+1)y=7 are perpendicular

Answer 1

The value of k is -3/7

Answer 2

you good?

Answer 3

how were you able to reach that answer..?

Answer 4

y = (k/3)x - 4/3

-(k + 1)y = -4x + 7
y = [4/(k + 1)]*x - [7/(k + 1)]

to be perpendicular the slopes must be negative reciprocals, then:

4/(k + 1) = -3/k
4k = -3k - 3
7k = -3
k = -3/7

Answer 5

ohmygod thankyou so much!
uhm uhm uhm is it possible for you to help me with another question?

Answer 6

maybe

Answer 7

I have to go soon because I'm on a tight schedule to finish my course.

Answer 8

Find a,b if the line ax+by+3=0 passes through (0,1) and is parallel to 2x-3y+5=0

only help if you have time !

Answer 9

i have to go

Answer 10

be back ltr

Answer 11

okay !

Answer 12

\[ax+by+3=0\]
Sub x = 0, y = 1 into the above equation:
\[b+3=0\]
\[b=-3\]
For the line
\[2x-3y+5=0\]
can be re-written as
\[y=\frac{ 2 }{ 3}x+\frac{ 5 }{ 3 }\]
Sub b = -3 back into ax + by + 3 = 0
\[-3y=-ax-3\]
\[y=\frac{ a }{ 3 }x+1\]
Since the two lines are parallel, their gradients are equal.
\[\frac{ a }{3 }=\frac{ 2 }{ 3 }\]
By inspection of numerators,
\[a=2\]