The area of the triangle with vertices (a, b) | vedclass

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(C) Given vertices are $(a, b)$,$(x_1, y_1) = (ar, bs)$,and $(x_2, y_2) = (ar^2, bs^2)$.The area of the triangle is given by the determinant formula:$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given vertices are $(a, b)$,$(x_1, y_1) = (ar, bs)$,and $(x_2, y_2) = (ar^2, bs^2)$.The area of the triangle is given by the determinant formula:$\Delta = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$Substituting the values:$\Delta = \frac{1}{2} |a(bs - bs^2) + ar(bs^2 - b) + ar^2(b - bs)|$Factoring out $ab$:$\Delta = \frac{1}{2} ab |(s - s^2) + r(s^2 - 1) + r^2(1 - s)|$$= \frac{1}{2} ab |s(1 - s) + r(s - 1)(s + 1) - r^2(s - 1)|$$= \frac{1}{2} ab |(s - 1) [-s + r(s + 1) - r^2]|$$= \frac{1}{2} ab |(s - 1) [-s + rs + r - r^2]|$$= \frac{1}{2} ab |(s - 1) [r(s - r) - (s - r)]|$$= \frac{1}{2} ab |(s - 1)(s - r)(r - 1)|$$= \frac{1}{2} ab (r - 1)(s - 1)(s - r)$.

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 $s$,is given by

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