Obtain the condition for the points (a,0), (0, b) and (1, 1) to be collinear.

Answer

To determine the condition for the points \( (a, 0) \), \( (0, b) \), and \( (1, 1) \) to be collinear, we can use the determinant method. The points are collinear if the area of the triangle formed by them is zero. The area can be calculated using the following determinant:

\[ \text{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]

Substituting the coordinates of the points \( (a, 0) \), \( (0, b) \), and \( (1, 1) \):

\[ \text{Area} = \frac{1}{2} \begin{vmatrix} a & 0 & 1 \\ 0 & b & 1 \\ 1 & 1 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ = \frac{1}{2} \left( a \begin{vmatrix} b & 1 \\ 1 & 1 \end{vmatrix} - 0 + 1 \begin{vmatrix} 0 & b \\ 1 & 1 \end{vmatrix} \right) \]

Calculating the smaller 2x2 determinants:

1. \( \begin{vmatrix} b & 1 \\ 1 & 1 \end{vmatrix} = b \cdot 1 - 1 \cdot 1 = b - 1 \)
2. \( \begin{vmatrix} 0 & b \\ 1 & 1 \end{vmatrix} = 0 \cdot 1 - b \cdot 1 = -b \)

Substituting back into the area formula:

\[ = \frac{1}{2} \left( a(b - 1) - b \right) \]

Setting the area to zero for collinearity:

\[ \frac{1}{2} \left( a(b - 1) - b \right) = 0 \]

This simplifies to:

\[ a(b - 1) - b = 0 \]

Rearranging gives:

\[ ab - a - b = 0 \]

Factoring out gives:

\[ ab - a - b + 1 = 1 \]

Thus, we can rewrite this as:

\[ ab = a + b \]

This is the condition for the points \( (a, 0) \), \( (0, b) \), and \( (1, 1) \) to be collinear.