Question

The points \( (K, 2 - 2K), (-K + 1, 2K) \) and \( (-4 - K, 6 - 2K) \) are collinear if: 

(A) \( K = \frac{1}{2} \) 
(B) \( K = -\frac{1}{2} \) 
(C) \( K = \frac{3}{2} \) 
(D) \( K = -1 \) 
(E) \( K = 1 \) 
 

• (A) and (D) only
• (A) and (E) only
• (B) and (D) only
• (D) only

Hint:

To check if points are collinear, use the area of the triangle formed by them and set it equal to zero.

Answer 1

The Correct Option is A

Step 1: Condition for collinearity of points.
For the points to be collinear, the area of the triangle formed by them should be zero. The area \( A \) of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is given by the formula: \[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] Substitute the coordinates of the points into the formula and solve for \( K \). After simplifying, we find that \( K = \frac{1}{2} \) and \( K = -1 \) satisfy the condition. Therefore, the correct answer is 1. (A) and (D) only.