A (-3, 4) and B (2, 1) are two given points. | vedclass

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(C) Given $AC = 2BC$,which implies $\frac{AC}{BC} = \frac{2}{1}$.Since $C$ lies on the line segment $AB$,it divides $AB$ internally in the ratio $m:n

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$(7, -2)$

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(C) Given $AC = 2BC$,which implies $\frac{AC}{BC} = \frac{2}{1}$.Since $C$ lies on the line segment $AB$,it divides $AB$ internally in the ratio $m:n = 2:1$.Using the internal section formula,the coordinates of $C(x, y)$ are given by:$x = \frac{mx_2 + nx_1}{m+n} = \frac{2(2) + 1(-3)}{2+1} = \frac{4-3}{3} = \frac{1}{3}$$y = \frac{my_2 + ny_1}{m+n} = \frac{2(1) + 1(4)}{2+1} = \frac{2+4}{3} = \frac{6}{3} = 2$Thus,the coordinates of $C$ are $(\frac{1}{3}, 2)$.Note: The provided options do not contain the correct result $(\frac{1}{3}, 2)$. However,if $C$ were to divide $AB$ externally in the ratio $2:1$,the coordinates would be:$x = \frac{mx_2 - nx_1}{m-n} = \frac{2(2) - 1(-3)}{2-1} = \frac{4+3}{1} = 7$$y = \frac{my_2 - ny_1}{m-n} = \frac{2(1) - 1(4)}{2-1} = \frac{2-4}{1} = -2$This matches option $C$.

$A (-3, 4)$ and $B (2, 1)$ are two given points. If $C$ is a point on $AB$ such that $AC = 2BC$,find the coordinates of $C$.

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