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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 the area of the triangle with vertices $(a, b), (x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is

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$|a b (r^{2} - 1)|$

$a b r s$

$a b (r^{2} - s^{2})$

$a b (s^{2} - 1)$

answer is A.

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Detailed Solution

Area of the triangle

$\begin{array}{l} = \frac{1}{2} |\begin{array}{ccc} a & b & 1 \\ a r & b s & 1 \\ a r^{2} & b s^{2} & 1 \end{array}| |= \frac{1}{2}| a b (r - 1) (s - 1) (s - r) ∣ \\ = | a b (r - 1) (r + 1) | = |a b (r^{2} - 1)| \end{array}$

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