if `(a,b) , (c,d) , (e,f)`are the vertices of a triangle such that `a, c, e` are in `GP`. with common ratio `r` and `b, d, f` are in `GP` with common ratio `s`, then the area of the triangle is

Options

A. `1/2|ab(r+1)(s+2)(s+r)|`
B. `1/2|ab(r-1)(s-1)(s-r)|`
C. `1/2|ab(r-1)(s+1) (s-r)|`
D. `1/2|ab(r+1)(s+1)(s-r)|`

Answer

To find the area of the triangle formed by the vertices \((a, b)\), \((c, d)\), and \((e, f)\) where \(a, c, e\) are in geometric progression (GP) with common ratio \(r\) and \(b, d, f\) are in GP with common ratio \(s\), we can follow these steps:

### Step 1: Define the coordinates based on the GP

Since \(a, c, e\) are in GP with common ratio \(r\):
- Let \(a\) be the first term.
- Then \(c = ar\) (the second term).
- And \(e = ar^2\) (the third term).

Similarly, since \(b, d, f\) are in GP with common ratio \(s\):
- Let \(b\) be the first term.
- Then \(d = bs\) (the second term).
- And \(f = bs^2\) (the third term).

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

The coordinates of the vertices of the triangle are:
- Vertex A: \((a, b)\)
- Vertex B: \((ar, bs)\)
- Vertex C: \((ar^2, bs^2)\)

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

The area \(A\) of a triangle given vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\) can be calculated using the determinant formula:

\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the coordinates of 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 the terms:
- The first term: \(a(bs - bs^2) = ab(s - s^2)\)
- The second term: \(ar(bs^2 - b) = ar(b(s^2 - 1))\)
- The third term: \(ar^2(b - bs) = ar^2(b(1 - s))\)

Combining these, we get:

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

### Step 5: Factor out common terms

Factoring out common terms from the expression, we can simplify further.

### Step 6: Final area expression

After performing the necessary algebraic manipulations, we arrive at the final expression for the area of the triangle:

\[ A = \frac{1}{2} ab (r - s)(1 - s)(1 - r) \]

### Conclusion

Thus, the area of the triangle formed by the vertices \((a, b)\), \((ar, bs)\), and \((ar^2, bs^2)\) is:

\[ A = \frac{1}{2} ab (r - s)(1 - s)(1 - r) \]