Question

the value of k for which the point (k, 2-2k), (-k, +1, 2k) and (-4, 2-2k) are collinear is​

Answer 1

Step-by-step explanation:

please post a picture of it

••••I hope this is wrong●●●●

Answer 2

Step-by-step explanation:

Consider the given points.

(−k+1,2k),(k,2−2k),(−4−k,6−2k)

Since, these points are collinear.

Therefore the area of triangle formed by the triangle formed by the points will be zero.

Therefore,

(−k+1)(2−2k−6+2k)+k(6−2k−2k)+(−4−k)(2k−2+2k)=0

(−k+1)(−4)+k(6−4k)+(−4−k)(4k−2)=0

4k−4+6k−4k

2

−16k+8−4k

2

+2k=0

−2k

2

−k+1=0

2k

2

+k−1=0

2k

2

+2k−k−1=0

2k(k+1)−1(k+1)=0

(k+1)(2k−1)=0

k=−1 or k=

2

1

Hence, this is the answer.