2 2 k 2 The value of TH) is 2 k 1
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Answer:
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
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