Question

Using determinants, find the value of \( \lambda \) if the points \( (1, -5) \), \( (-4, 5) \), and \( (\lambda, 7) \) are collinear.

Hint:

For collinearity using determinants: If determinant of 3×3 matrix = 0, points lie on the same straight line.

Answer 1

Concept: Three points are collinear if the area of the triangle formed by them is zero. Using determinant method: \[ \left| \begin{matrix} x_1 & y_1 & 1
x_2 & y_2 & 1
x_3 & y_3 & 1 \end{matrix} \right| = 0 \] Step 1: Substitute the given points.
\[ \left| \begin{matrix} 1 & -5 & 1
-4 & 5 & 1
\lambda & 7 & 1 \end{matrix} \right| = 0 \]
Step 2: Expand the determinant.
\[ = 1 \left| \begin{matrix} 5 & 1
7 & 1 \end{matrix} \right| - (-5) \left| \begin{matrix} -4 & 1
\lambda & 1 \end{matrix} \right| + 1 \left| \begin{matrix} -4 & 5
\lambda & 7 \end{matrix} \right| \] \[ = 1(5 \cdot 1 - 7 \cdot 1) + 5((-4 \cdot 1 - \lambda \cdot 1)) + 1((-4 \cdot 7 - 5\lambda)) \] \[ = (5 - 7) + 5(-4 - \lambda) + (-28 - 5\lambda) \] \[ = -2 -20 -5\lambda -28 -5\lambda \] \[ = -50 -10\lambda \]
Step 3: Set equal to zero.
\[ -50 -10\lambda = 0 \] \[ -10\lambda = 50 \] \[ \lambda = -5 \] Correction:
Rechecking calculation carefully: \[ = -2 + (-20 -5\lambda) + (-28 -5\lambda) \] \[ = -2 -20 -28 -10\lambda \] \[ = -50 -10\lambda \] \[ -50 -10\lambda = 0 \Rightarrow \lambda = -5 \] Conclusion:
The correct value of \( \lambda \) is: \[ \lambda = -5 \]