Question

find the value of k of the following points are on the straight line. (k, 2-2k), (1-k,2k) and (-4,6-2k).​

Answer 1

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−4k2−16k+8−4k2+2k=0

−2k2−k+1=0

2k2+k−1=0

2k2+2k−k−1=0

2k(k+1)−1(k+1)=0

(k+1)(2k−1)=0

k=−1 or k=21

Answer 2

Answer:

$k \: = \: - 1 \: or \: k \: = \: \frac{1}{2}$

hope it's help you