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 $r$.Therefore,we can write: ${x_2} = r{x_1},

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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 $r$.Therefore,we can write: ${x_2} = r{x_1}, {x_3} = {r^2}{x_1}$ and ${y_2} = r{y_1}, {y_3} = {r^2}{y_1}$.To check if the points $({x_1}, {y_1}), ({x_2}, {y_2}), ({x_3}, {y_3})$ are collinear,we calculate the area of the triangle formed by these points: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:Area $= \frac{1}{2} |x_1(ry_1 - r^2y_1) + rx_1(r^2y_1 - y_1) + r^2x_1(y_1 - ry_1)|$Area $= \frac{1}{2} |x_1y_1(r - r^2) + x_1y_1(r^3 - r) + x_1y_1(r^2 - r^3)|$Area $= \frac{1}{2} |x_1y_1(r - r^2 + r^3 - r + r^2 - r^3)| = 0$.Since the area of the triangle is $0$,the points are collinear and 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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