Question

Find the value of k for which the points A(k+1,2k), B(3k,2k+3) and C(5k-1,5k) are collinear​

Answer 1

Given :

$\sf A = \: (k + 1, 2k)$

$\sf B = \: (3k, 2k + 3)$

$\sf C = \: (5k - 1, 5k)$

To Find :

The value of k.

Solution :

According to the question,

The points are collinear then,

$\sf \dashrightarrow {x}_{1} ({y}_{2} - {y}_{3}) + {x}_{2} ({y}_{3} - {y}_{1}) + {x}_{3} ({y}_{1} - {y}_{2}) = 0$

$\sf \dashrightarrow ({x}_{1} = k + 1, {y}_{1} = 2k)$

$\sf \dashrightarrow ({x}_{2} = 3k, {y}_{2} = 2k + 3$

$\sf \dashrightarrow {x}_{3} = 5k - 1, {y}_{3} = 5k$

$\sf \dashrightarrow (k + 1)(2k + 3 - 5k) + 3k (5k - 2k) + (5k - 1) [2k - (2k + 3)] = 0$

$\sf \dashrightarrow (k + 1)(3 - 3k) + 3k \times 3k + (5k - 1)\times (-3) = 0$

$\sf \dashrightarrow 3 - {3k}^{2} + {9k}^{2} - 15k + 3 = 0$

$\sf \dashrightarrow {6k}^{2} - 15k + 6 = 0$

$\sf \dashrightarrow {2k}^{2} - 5k + 2 = 0$

$\sf \dashrightarrow {2k}^{2} - 4k - k + 2 = 0$

$\sf \dashrightarrow 2k(k - 2) - (k - 2) = 0$

$\sf \dashrightarrow (k - 2)(2k - 1) = 0$

$\sf \dashrightarrow k - 2 = 0 \: \: or \: \: 2k - 1 = 0$

$\sf \dashrightarrow k = 2 \: \: or \: \: k = \dfrac{1}{2}$

Hence, k = 2 or k = ½