Published by:
CGP EDU Academic Team
Published on: August 14, 2026
If the points (k, 2 – 2k), (1 – k, 2k) and (–k –4, 6 – 2k) are collinear, then possible values of k are
A
– 
, 1
, 1
B
, – 1
C
1, 2
D
1, 3
Text Solution
Verified by ExpertsThe correct answer is:
B
Since A, B, C are collinear 
Slope of AB = Slope of BC A(k, 2 – 2k) B(1 – k, 2k) C(– k– 4, 6 – 2k)
= 
⇒ 
= 
⇒ 10 – 20k = (4k – 6)(2k – 1)
⇒ (4k – 6)(2k – 1) + 10(2k – 1) = 0
⇒ k = 
, – 1
──────────────────────────────────────────────────────────────────────────────────────────
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
The vertices of a triangle OBC are O(0,0) B(–3,–1) and C(–1,–3). Find the equation of line parallel…
A variable line, drawn through the point of intersection of the straight lines = 1 and = 1, meet…
From the vertices A, B, C of a triangle ABC, perpendiculars AD, BE, CF are drawn to any straight li…
A triangle is formed by the lines whose equations are AB : x + y – 5 = 0, BC : x + 7y – 7 = 0 and …
The coordinates of the feet of from the vertices of a Δ on the opposite sides are (20, 25), (8, 16…
Let P is any point inside the triangle ABC of side lengths 6, 5, 5 units and p 1 , p 2 , p 3 be the…