Answer: \(AC:AB=1:3\).

The vector from \(A\) to \(B\) is

\(\overrightarrow{AB}=\binom{7}{4}-\binom{1}{7}=\binom{6}{-3}.\)

A point on the line segment \(AB\) can be written as

\(\overrightarrow{OA}+t\overrightarrow{AB}=\binom{1}{7}+t\binom{6}{-3}=\binom{1+6t}{7-3t},\)

where \(0\leqslant t\leqslant1\). The value of \(t\) represents the fraction of the way from \(A\) to \(B\).

Since \(C\) has position vector

\(\overrightarrow{OC}=k\binom12=\binom{k}{2k},\)

we equate components:

\(k=1+6t,\qquad 2k=7-3t.\)

Substitute \(k=1+6t\) into the second equation:

\(2(1+6t)=7-3t.\)

Thus

\(2+12t=7-3t,\)

so

\(15t=5,\qquad t=\dfrac13.\)

Therefore \(C\) is one third of the way from \(A\) to \(B\). Hence

\(AC:AB=\dfrac13:1=1:3.\)