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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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ab(r2-1)

ab(r2-s2)

ab(s2-1)

abrs

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detailed solution

Correct option is A

Area of the triangle=12 ab1arbs1ar2bs21=12ab(r−1)(s−1)(s−r)∣ =|ab(r−1)(r+1)|=abr2−1

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If a, x1, x2 are in G.P. with common ratio r, and b, y1, y2 are in G.P. with common ratio s where s – r = 2, then the area of the triangle with vertices (a, b), (x1, y1) and (x2, y2) is

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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