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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) , $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ is

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

$| \frac{a b (r - s) (s - 1) (r - 1)}{2} |$

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

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

answer is D.

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Detailed Solution

$= \frac{1}{2} |\begin{matrix} a & b & 1 \\ a r & b s & 1 \\ a r^{2} & b s^{2} & 1 \end{matrix}|$

$= |\frac{a b}{2} |\begin{matrix} 1 & 1 & 1 \\ r & s & 1 \\ r^{2} & s^{2} & 1 \end{matrix}||$

$= |\frac{a b}{2} (r - s) (s - 1) (1 - r)|$

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