Question

if the points A(K+1,2K), B(3K,2K+3),and C(5K-1,5K) are collinear,then find the value of k

Answer 1

Sol:

(i) Vertices of the triangle are P(a,b),Q(b,c) and R(c,a)
Centroid = [(a+b+c)/3, (a+b+c)/3]
⇒ [(a+b+c)/3, (a+b+c)/3] = (0, 0)
⇒ (a+b+c)/3 = 0
⇒ (a+b+c) = 0

(ii) The points A(k+1, 2k), B(3k, 2k+3) and C(5k-1, 5k) are collinear.

Area of the triangle formed by A, B and C
= 1/2 {(k+1)[(2k+3) - (5k)] + (3k) [(5k) - (k+1)] + (5k-1)[(2k) - (2k+3)]} = 0
= 1/2 {[(k + 1)(-3k + 3)] + [(3k)(4k - 1)] +(5k - 1)(3)]} = 0
= 1/2 {(-3k2 + 3) + (12k2 - 3k) + (15k - 3)} = 0
= 1/2 {9k2 + 12k} = 0
= 9k + 12 = 0
k = -12/9
k = -4/3