If {x_1}, {x_2}, {x_3} and {y_1}, {y_2}, {y_3 | vedclass

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(A) Let the common ratio be $r$. Since ${x_1}, {x_2}, {x_3}$ are in $G$.$P$.,we have ${x_2} = {x_1}r$ and ${x_3} = {x_1}r^2$.Similarly,since ${y_1}, {

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Lie on a straight line

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(A) Let the common ratio be $r$. Since ${x_1}, {x_2}, {x_3}$ are in $G$.$P$.,we have ${x_2} = {x_1}r$ and ${x_3} = {x_1}r^2$.Similarly,since ${y_1}, {y_2}, {y_3}$ are in $G$.$P$. with the same common ratio $r$,we have ${y_2} = {y_1}r$ and ${y_3} = {y_1}r^2$.The points are $P_1 = ({x_1}, {y_1})$,$P_2 = ({x_1}r, {y_1}r)$,and $P_3 = ({x_1}r^2, {y_1}r^2)$.Notice that for all points,the ratio of the $y$-coordinate to the $x$-coordinate is constant: $\frac{{y_1}}{{x_1}} = \frac{{y_1}r}{{x_1}r} = \frac{{y_1}r^2}{{x_1}r^2} = m$ (where $m = \frac{{y_1}}{{x_1}}$).This implies that all three points satisfy the linear equation $y = mx$,which represents a straight line passing through the origin.

If ${x_1}, {x_2}, {x_3}$ and ${y_1}, {y_2}, {y_3}$ are both 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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