Find the condition for are point (a,0), (h,k), (o,b) where `ab ne0` tobe collinear.

Answer

To determine the condition for the points (a, 0), (h, k), and (0, b) to be collinear, we can use the concept of the area of a triangle formed by these three points. If the area is zero, then the points are collinear.

### Step-by-Step Solution:

1. **Set Up the Points**:
We have three points:
- Point 1: \( P_1(a, 0) \)
- Point 2: \( P_2(h, k) \)
- Point 3: \( P_3(0, b) \)

2. **Use the Area Formula**:
The area \( A \) of the triangle formed by these three points can be calculated using the determinant:
\[ A = \frac{1}{2} \left| \begin{vmatrix} a & 0 & 1 \\ h & k & 1 \\ 0 & b & 1 \end{vmatrix} \right| \]
For the points to be collinear, the area must be zero:
\[ \begin{vmatrix} a & 0 & 1 \\ h & k & 1 \\ 0 & b & 1 \end{vmatrix} = 0 \]

3. **Calculate the Determinant**:
Expanding the determinant:
\[ = a \begin{vmatrix} k & 1 \\ b & 1 \end{vmatrix} - 0 + 1 \begin{vmatrix} h & k \\ 0 & b \end{vmatrix} \]
This simplifies to:
\[ = a(k \cdot 1 - b \cdot 1) + h \cdot b \]
\[ = ak - ab + hb \]

4. **Set the Determinant to Zero**:
For collinearity, we set the expression equal to zero:
\[ ak - ab + hb = 0 \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ ak + hb = ab \]

6. **Divide by ab**:
Since \( ab \neq 0 \), we can divide the entire equation by \( ab \):
\[ \frac{ak}{ab} + \frac{hb}{ab} = 1 \]
This simplifies to:
\[ \frac{k}{b} + \frac{h}{a} = 1 \]

### Final Condition:
The condition for the points (a, 0), (h, k), and (0, b) to be collinear is:
\[ \frac{h}{a} + \frac{k}{b} = 1 \]