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(A) Given $AB = 3AM$. Since $A-M-B$,we have $AB = AM + MB$. Substituting $AB$,we get $AM + MB = 3AM$,which implies $MB = 2AM$. Thus,the ratio $AM : MB

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$\left( \frac{8}{3}, \frac{10}{3} \right)$

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(A) Given $AB = 3AM$. Since $A-M-B$,we have $AB = AM + MB$. Substituting $AB$,we get $AM + MB = 3AM$,which implies $MB = 2AM$. Thus,the ratio $AM : MB = 1 : 2$. Using the section formula for a point $M(x, y)$ dividing $AB$ in the ratio $m:n = 1:2$,where $A(x_1, y_1) = (2, 4)$ and $B(x_2, y_2) = (4, 2)$: $x = \frac{mx_2 + nx_1}{m+n} = \frac{1(4) + 2(2)}{1+2} = \frac{4+4}{3} = \frac{8}{3}$. $y = \frac{my_2 + ny_1}{m+n} = \frac{1(2) + 2(4)}{1+2} = \frac{2+8}{3} = \frac{10}{3}$. Therefore,the coordinates of $M$ are $\left( \frac{8}{3}, \frac{10}{3} \right)$.

If coordinates of the points $A$ and $B$ are $(2, 4)$ and $(4, 2)$ respectively and point $M$ is such that $A-M-B$ and $AB = 3AM$,then the coordinates of $M$ are

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