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
View Solution FreeMore Point Questions
If the points (1, 1) (-1, -1) and (- √3 , k) are vertices of an equilateral triangle then the value of k will be -...If the point (a, 2) lies between the lines x - y - 1 = 0 and 2(x - y) - 5 = 0, then the set of values of a is -...The distance between feet of perpendiculars drawn from a point (−3, 4) on both axes is -...Orthocentre of a triangle whose vertex are (8, −2), (2, − 2) and (8, 6) is -...The x coordinate of the incentre of the triangle where the mid point of the sides are (0, 1), (1, 1) and (1, 0) is -...