Question

For what value of \( K \), points \( (K, -1) \), \( (2, 1) \), and \( (4, 5) \) will be on the same line?

Hint:

For collinearity, use the area method or check if the determinant of the points equals zero.

Answer 1

For three points to be collinear, the area of the triangle formed by them must be zero. The area of the triangle formed by points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the formula: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] For collinearity, the area must be zero, so: \[ 0 = \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] Substitute the points \( (K, -1) \), \( (2, 1) \), and \( (4, 5) \): \[ 0 = \left| K(1 - 5) + 2(5 - (-1)) + 4((-1) - 1) \right| \] Simplify the terms: \[ 0 = \left| K(-4) + 2(6) + 4(-2) \right| \] \[ 0 = \left| -4K + 12 - 8 \right| \] \[ 0 = \left| -4K + 4 \right| \] \[ -4K + 4 = 0 \] Solve for \( K \): \[ -4K = -4 \quad \Rightarrow \quad K = 1 \]
Conclusion:
For the points to be on the same line, the value of \( K \) must be \( 1 \).