If {x_1}, {x_2}, {x_3} as well as {y_1}, {y_2 | vedclass

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(A) Given that ${x_1}, {x_2}, {x_3}$ and ${y_1}, {y_2}, {y_3}$ are in $G$.$P$. with the same common ratio $r$,we have:${x_2} = r{x_1}, {x_3} = {r^2}{x

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Lie on a straight line

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(A) Given that ${x_1}, {x_2}, {x_3}$ and ${y_1}, {y_2}, {y_3}$ are in $G$.$P$. with the same common ratio $r$,we have:${x_2} = r{x_1}, {x_3} = {r^2}{x_1}$${y_2} = r{y_1}, {y_3} = {r^2}{y_1}$To check if the points are collinear,we calculate the area of the triangle formed by these points:$	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$Substituting the values:$	ext{Area} = \frac{1}{2} |x_1(ry_1 - r^2y_1) + rx_1(r^2y_1 - y_1) + r^2x_1(y_1 - ry_1)|$$	ext{Area} = \frac{1}{2} |x_1y_1(r - r^2) + x_1y_1(r^3 - r) + x_1y_1(r^2 - r^3)|$$	ext{Area} = \frac{1}{2} |x_1y_1(r - r^2 + r^3 - r + r^2 - r^3)| = 0$Since the area is $0$,the points lie on a straight line.

If ${x_1}, {x_2}, {x_3}$ as well as ${y_1}, {y_2}, {y_3}$ are in $G$.$P$. with the same common ratio,then the points $({x_1}, {y_1}), ({x_2}, {y_2})$ and $({x_3}, {y_3})$:

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