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

Answer

To find the area of the triangle with vertices \((a,b)\), \((x_1,y_1)\), and \((x_2,y_2)\) given that \(a, x_1, x_2\) are in G.P. with common ratio \(r\) and \(b, y_1, y_2\) are in G.P. with common ratio \(s\) where \(s - r = 2\), we can follow these steps:

### Step 1: Express \(x_1\), \(x_2\), \(y_1\), and \(y_2\)

Since \(a, x_1, x_2\) are in G.P. with common ratio \(r\):
- \(x_1 = ar\)
- \(x_2 = ar^2\)

Since \(b, y_1, y_2\) are in G.P. with common ratio \(s\):
- \(y_1 = bs\)
- \(y_2 = bs^2\)

### Step 2: Write the vertices of the triangle

The vertices of the triangle can be written as:
- Vertex 1: \((a, b)\)
- Vertex 2: \((ar, bs)\)
- Vertex 3: \((ar^2, bs^2)\)

### Step 3: Use the formula for the area of a triangle

The area \(A\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the vertices:
\[ A = \frac{1}{2} \left| a(bs - bs^2) + ar(bs^2 - b) + ar^2(b - bs) \right| \]

### Step 4: Simplify the expression

Calculating each term:
1. \(a(bs - bs^2) = ab(s - s^2)\)
2. \(ar(bs^2 - b) = ar(b(s^2 - 1))\)
3. \(ar^2(b - bs) = ar^2(b(1 - s))\)

Now substituting back:
\[ A = \frac{1}{2} \left| ab(s - s^2) + ar(b(s^2 - 1)) + ar^2(b(1 - s)) \right| \]

### Step 5: Factor out common terms

We can factor out \(ab\):
\[ A = \frac{1}{2} \left| ab \left( (s - s^2) + r(s^2 - 1) + r^2(1 - s) \right) \right| \]

### Step 6: Substitute \(s - r = 2\)

From the problem, we know \(s - r = 2\). Thus, we can express \(s\) in terms of \(r\):
\[ s = r + 2 \]

Now substitute \(s\) back into the area expression and simplify further.

### Final Expression for Area

After substituting and simplifying, we will arrive at:
\[ A = \frac{1}{2} \left| ab \cdot \text{(some expression in terms of } r \text{ and } b \text{)} \right| \]