If `a, x_(1), x_(2)` are in G.P. with common ration r, and `b, y_(1), y_(2)` are in G.P. with common ratio s where `s-r=2`, then the area of the triangle with vertices `(a, b), (x_(1), y_(1))` and `(x_(2),y_(2))` is

Options

A. `|ab(r^(2)-1)|`
B. `ab(r^(2)-s^(2)`
C. `ab(s^(2)-1)`
D. abrs

Answer

To solve the problem, we need to find the area of a triangle with vertices at points \((a, b)\), \((x_1, y_1)\), and \((x_2, y_2)\) given that \(a, x_1, x_2\) are in geometric progression (G.P.) with common ratio \(r\) and \(b, y_1, y_2\) are in G.P. with common ratio \(s\), where \(s - r = 2\).

### Step-by-Step Solution:

1. **Identify the terms in G.P.**:
- Since \(a, x_1, x_2\) are in G.P. with common ratio \(r\):
\[ x_1 = ar \quad \text{and} \quad x_2 = ar^2 \]
- Since \(b, y_1, y_2\) are in G.P. with common ratio \(s\):
\[ y_1 = bs \quad \text{and} \quad y_2 = bs^2 \]

2. **Use the formula for the area of a triangle**:
The area \(A\) of a triangle with vertices at \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For our points \((a, b)\), \((ar, bs)\), and \((ar^2, bs^2)\), we can substitute:
\[ A = \frac{1}{2} \left| a(bs^2 - b) + ar(bs - b) + ar^2(b - bs) \right| \]

3. **Simplify the expression**:
- First, factor out \(b\):
\[ A = \frac{1}{2} \left| a(b(s^2 - 1)) + ar(b(s - 1)) + ar^2(b(1 - s)) \right| \]
- This can be simplified to:
\[ A = \frac{1}{2} b \left| a(s^2 - 1) + ar(s - 1) + ar^2(1 - s) \right| \]

4. **Substitute \(s = r + 2\)**:
- Now, substitute \(s\) in the area expression:
\[ A = \frac{1}{2} b \left| a((r + 2)^2 - 1) + ar((r + 2) - 1) + ar^2(1 - (r + 2)) \right| \]
- Expanding this gives:
\[ A = \frac{1}{2} b \left| a(r^2 + 4r + 4 - 1) + ar(r + 1) + ar^2(-r - 1) \right| \]
- Simplifying further:
\[ A = \frac{1}{2} b \left| a(r^2 + 4r + 3) + ar^2 - ar^2 - ar \right| \]
- This simplifies to:
\[ A = \frac{1}{2} b \left| a(r^2 + 4r + 3 - r) \right| = \frac{1}{2} b \left| a(3 + 3r) \right| \]

5. **Final expression for area**:
- Therefore, the area can be expressed as:
\[ A = \frac{1}{2} ab \left| 3 + 3r \right| = \frac{3}{2} ab |1 + r| \]

### Final Answer:
Thus, the area of the triangle with vertices \((a, b)\), \((x_1, y_1)\), and \((x_2, y_2)\) is:
\[ \frac{3}{2} ab |1 + r| \]