Question

3 Find K if the points that
(K, 2-2K) (-k+1, 2k)
(-4-K, 6-2K) are collinear​

Answer 1

Answer:

Step-by-step explanation:

There is a restrictions for 3 points to be colinear and that is

1/2[X1(Y2-Y3)+X2(Y3-Y1)+X3(Y1-Y2)] = 0

Here X1= -K+1, Y1= 2K, X2=K, Y2= 2–2K, X3= -4-K, Y3= 6–2K

1/2[-K+1(2–2k -6+2k) + K(6–2k -2k) -4-K(2k -2+2k)] =0

1/2[-K+1(-4) +K (6- 4K) -4-k(4K-2)]=0

1/2[4K -4 +6K -4k^2 -16K +8 -4k^2 +2k)] =0

1/2(-8K^2 -4K +4)=0

-8K^2 -4K +4=0

-8K^2 -8K +4K +4 =0

-8K( K + 1 ) +4( K + 1)=0

( K+ 1) ( 4 -8k )=0

K+ 1= 0

K = -1

4 -8k =0

K= 4/8

K = 1/2

So the value of K is -1 and 1/2

hope it helps

:)

Answer 2

Answer:

its 37

Step-by-step explanation:

did it in khan