Question

For what value of k, the points (k,-1),(5,7)and (8,11)are collinear?

Answer 1

Answer:

For the points to be collinear,

x1(y2-y3) + x2(y3-y1) + x3(y1-y2)=0... (1)

now the points are A(1,5) B(K,1) and C(4,11) i.e.

x1=1 y1=5

x2=K y2=1

x3=4 y3=11

put these values in equation 1..

1(1-11) + K(11-5) + 4(5-1) =0

-10+ 6K +16 =0

6K+6=0.... 6K=-6

i.e. K=-1...

Answer 2

Answer:

Hope it's may help you

please mark me as a brainalist