Think!!! Do without Graph!!!
Geometry
Level
pending
If the points (k,2-2k) (-k+1,2k) (-4-k,6-2k) are collinear .Find the possible value of k _ _ ?
2,1/3
3,1/2
1/3,0
1/3,-2
1/2 , -1
No real Values
3,-1
1,1/2
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1 solution
If the points are collinear then the area of triangle formed by joining the points will be equal to ZERO
using the formula:
Area=(1/2) [ k(2k-(6-2k)) + (1-k)(6-2k-(2-2k)) + (-4-k)(2-2k-(2k)) ]=0 solving the following we get 2k(k+1)-1(k+1)=0 => 2k-1=0 (or)k+1=0 => k=1/2 (or)k=-1