Coordinate Geometry Question 154
Question: If the points $ (k,2-2k) $ , $ (1-k,2k) $ and $ (-k-4,6-2k) $ be collinear, then the possible values of k are
[AMU 1978; RPET 1997]
Options:
A) $ \frac{1}{2},-1 $
B) $ 1,-\frac{1}{2} $
C) $ 1,-2 $
D) $ 2,-1 $
Show Answer
Answer:
Correct Answer: A
Solution:
The points are collinear if the area of triangle formed by these three points is zero.
$ \Rightarrow \frac{1}{2}[k{2k-(6-2k)}+(1-k){(6-2k)-(2-2k)} $
$ +(-4-k){(2-2k)-2k}]=0 $ On simplification, we get $ k=-1 $ or $ \frac{1}{2} $ .
BETA
