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(C) Given that $x_1 = ar, x_2 = ar^2$ and $y_1 = bs, y_2 = bs^2$.The area of the triangle $\Delta$ is given by the determinant formula:$\Delta = \frac

Vedclass · Accepted Answer

$\frac{1}{2} ab (r - 1) (s - 1) (s - r)$

Vedclass · Answer

(C) Given that $x_1 = ar, x_2 = ar^2$ and $y_1 = bs, y_2 = bs^2$.The area of the triangle $\Delta$ is given by the determinant formula:$\Delta = \frac{1}{2} \left| \begin{matrix} a & b & 1 \ x_1 & y_1 & 1 \ x_2 & y_2 & 1 \end{matrix} ight| = \frac{1}{2} \left| \begin{matrix} a & b & 1 \ ar & bs & 1 \ ar^2 & bs^2 & 1 \end{matrix} ight|$Taking $a$ common from the first column and $b$ common from the second column:$\Delta = \frac{1}{2} ab \left| \begin{matrix} 1 & 1 & 1 \ r & s & 1 \ r^2 & s^2 & 1 \end{matrix} ight|$Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\Delta = \frac{1}{2} ab \left| \begin{matrix} 1 & 1 & 1 \ r-1 & s-1 & 0 \ r^2-1 & s^2-1 & 0 \end{matrix} ight|$Expanding along the third column:$\Delta = \frac{1}{2} ab \left| \begin{matrix} r-1 & s-1 \ (r-1)(r+1) & (s-1)(s+1) \end{matrix} ight|$$\Delta = \frac{1}{2} ab (r-1)(s-1) \left| \begin{matrix} 1 & 1 \ r+1 & s+1 \end{matrix} ight|$$\Delta = \frac{1}{2} ab (r-1)(s-1) (s+1 - r - 1) = \frac{1}{2} ab (r-1)(s-1)(s-r)$.

Find the area of the triangle with vertices $(a, b)$,$(x_1, y_1)$,and $(x_2, y_2)$,where $a, x_1, x_2$ are in $G.P.$ with common ratio $r$,and $b, y_1, y_2$ are in $G.P.$ with common ratio $s$.

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