If point \( (-1,+-1),(t, t-3) \) and \( (t,-6,3) \) are collinear, pund the value of \( t \)

To determine the value of \( t \) for which the points \((-1, -1)\), \((t, t-3)\), and \((t, -6)\) are collinear, we can use the concept of the area of the triangle formed by these points. The area is zero if the points are collinear. Using the determinant formula for the area of the triangle formed by three points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\): \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] Substituting in our points: \[ \text{Area} = \frac{1}{2} \left| -1((t-3) - (-6)) + t(-6 - (-1)) + t-3(-1 - (t-3)) \right| = 0 \] Calculating the terms inside the absolute value: 1. For \(-1((t-3) + 6) = -1(t + 3)\). 2. For \(t(-6 + 1) = -5t\). 3. For \(t-3(2 - t) = -t^2 + 5t - 6\). Combining all terms gives: \[ -t - 3 - 5t - t^2 + 5t - 6 = -t^2 - t - 9 = 0 \] This simplifies to: \[ t^2 + t + 9 = 0 \] Calculating the discriminant: \[ D = b^2 - 4ac = 1^2 - 4 \times 1 \times 9 = 1 - 36 = -35 \] Since the discriminant is negative, there are no real solutions for \( t \).