PointHard
Question
The points with the co-ordinates (2a, 3a), (3b, 2b) & (c, c) are collinear -
Options
A.for no value of a, b, c
B.for all values of a, b c
C.if a, c/5 , b are in H. P.
D.if a, 2/5 c , b are in H. P
Solution
ᐃ = 
= 0
(2a - c) (2b - c) - (3a - c) (3b - c) = 0
⇒ 4ab - 2ac - 2bc + c2 - (9ab - 3ac - 3bc + c2) = 0
⇒ ac + bc - 5ab = 0
⇒ 
∴ a,
, b are in H. P.
= 0(2a - c) (2b - c) - (3a - c) (3b - c) = 0
⇒ 4ab - 2ac - 2bc + c2 - (9ab - 3ac - 3bc + c2) = 0
⇒ ac + bc - 5ab = 0
⇒ ∴ a,
, b are in H. P.Create a free account to view solution
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