If P(a, 0), Q(0, b) and R(1, 1) are collinear. Then, find the value of `1/a + 1/b`.

Options

A. 2
B. 1
C. -1
D. 0

Answer

To solve the problem, we need to determine the value of \( \frac{1}{a} + \frac{1}{b} \) given that the points \( P(a, 0) \), \( Q(0, b) \), and \( R(1, 1) \) are collinear.

### Step-by-Step Solution:

1. **Identify the Coordinates**:
- The coordinates of the points are:
- \( P(a, 0) \) → \( (x_1, y_1) = (a, 0) \)
- \( Q(0, b) \) → \( (x_2, y_2) = (0, b) \)
- \( R(1, 1) \) → \( (x_3, y_3) = (1, 1) \)

2. **Use the Collinearity Condition**:
- The points are collinear if the area of the triangle formed by them is zero. The area can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
- Setting this area to zero gives us the equation for collinearity:
\[ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0 \]

3. **Substituting the Coordinates**:
- Substitute the coordinates into the collinearity equation:
\[ a(b - 1) + 0(1 - 0) + 1(0 - b) = 0 \]
- Simplifying this gives:
\[ a(b - 1) - b = 0 \]

4. **Rearranging the Equation**:
- Rearranging the equation:
\[ ab - a - b = 0 \]
- This can be rewritten as:
\[ ab - a - b + 1 = 1 \]
- Factoring gives:
\[ (a - 1)(b - 1) = 1 \]

5. **Finding \( \frac{1}{a} + \frac{1}{b} \)**:
- From the equation \( (a - 1)(b - 1) = 1 \), we can express \( a \) and \( b \):
\[ ab - a - b + 1 = 1 \implies ab = a + b \]
- Now, we can find \( \frac{1}{a} + \frac{1}{b} \):
\[ \frac{1}{a} + \frac{1}{b} = \frac{b + a}{ab} = \frac{a + b}{ab} \]
- Since \( ab = a + b \), we have:
\[ \frac{1}{a} + \frac{1}{b} = \frac{a + b}{a + b} = 1 \]

### Conclusion:
Thus, the value of \( \frac{1}{a} + \frac{1}{b} \) is \( 1 \).