Question

if points ( k , 2k ) ; (3k ,3k ) and ( 3 ,1 ) are collinear then find value of k.​

Answer 1

Answer:-

Given:-

(k , 2k) ; (3k , 3k) and (3 , 1) are collinear points i.e., they lie on the same line.

We know that;

If three points are collinear, when the area of the triangle formed by them is 0 square units.

Also;

$\sf \: Area \: of \: a \: triangle = \frac{1}{2} \begin{vmatrix} \sf x_1 - x_2& \sf \: x_1 - x_3 \\ \\ \sf \: y_1 - y_2& \sf \: y_1 - y_3 \end{vmatrix}$

Let;

• x₁ = k

• x₂ = 3k

• x₃ = 3

• y₁ = 2k

• y₂ = 3k

• y₃ = 1

According to the question;

$\implies \sf \: \frac{1}{2} \begin{vmatrix} \sf k - 3k& \sf \: k - 3 \\ \\ \sf \:2k - 3k & \sf \: 2k - 1 \: \end{vmatrix} = 0 \\ \\ \\ \implies \sf \:\begin{vmatrix} \sf - 2k& \sf \: k - 3 \\ \\ \sf \: - k & \sf \: 2k - 1 \: \end{vmatrix} = 0 \times 2 \\ \\ \\ \implies \sf \: |( - 2k)(2k - 1) - ( - k)(k - 3)| = 0 \\ \\ \\ \implies \sf | - 4 {k}^{2} + 2k - ( - {k}^{2} + 3k) | = 0 \\ \\ \\ \implies \sf \: | - 4 {k}^{2} + 2k + {k}^{2} - 3k | = 0 \\ \\ \\ \implies \sf \: | - 3 {k}^{2} - k| = 0 \\ \\ \\ \implies \sf \: 3 {k}^{2} + k= 0 \\ \\ \\ \implies \sf \: k(3k + 1) = 0 \\ \\ \\ \implies \sf \: k = 0 \: \: \: (or) \: \: 3k + 1 = 0 \\ \\ \\ \implies \boxed{\sf \:k = 0 \: \: (or) \: \: \frac{ - 1}{3}}$

∴ The value of k is - 1/3 or 0.