If three points (h, 0), (a, b) and (0, k) lies on a line, show that `(a)/(h) + (b)/(k) = 1`.

Answer

To show that if the points \((h, 0)\), \((a, b)\), and \((0, k)\) are collinear, then \(\frac{a}{h} + \frac{b}{k} = 1\), we can use the concept of the area of the triangle formed by these three points. If the area is zero, it indicates that the points are collinear.

### Step-by-Step Solution:

1. **Identify the Points**:
We have three points:
- \(P_1 = (h, 0)\)
- \(P_2 = (a, b)\)
- \(P_3 = (0, k)\)

2. **Area of Triangle Formula**:
The area \(A\) of a triangle formed by the points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For our points, this becomes:
\[ A = \frac{1}{2} \left| h(b - k) + a(k - 0) + 0(0 - b) \right| \]

3. **Substituting the Points**:
Substitute the coordinates into the area formula:
\[ A = \frac{1}{2} \left| h(b - k) + ak \right| \]

4. **Setting Area to Zero**:
Since the points are collinear, the area must be zero:
\[ \frac{1}{2} \left| h(b - k) + ak \right| = 0 \]
This implies:
\[ h(b - k) + ak = 0 \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ h(b - k) = -ak \]
Dividing both sides by \(hk\) (assuming \(h \neq 0\) and \(k \neq 0\)):
\[ \frac{b - k}{k} = -\frac{a}{h} \]

6. **Simplifying**:
This can be rewritten as:
\[ \frac{b}{k} - 1 = -\frac{a}{h} \]
Rearranging gives:
\[ \frac{b}{k} + \frac{a}{h} = 1 \]

7. **Final Result**:
Thus, we have shown that:
\[ \frac{a}{h} + \frac{b}{k} = 1 \]