There is no arrow straight from \(A\) to \(B\), so travel there through \(O\):

\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB}\)

\(\overrightarrow{AO}\) is \(\overrightarrow{OA}\) backwards, so \(\overrightarrow{AO} = -12\mathbf{a}\).

\(\overrightarrow{AB} = -12\mathbf{a} + 12\mathbf{b} = 12\mathbf{b} - 12\mathbf{a}\)

\(AP : PB = 3 : 1\) means \(AB\) is cut into \(4\) equal parts and \(AP\) is \(3\) of them, so \(\overrightarrow{AP} = \frac{ 3 }{ 4 }\overrightarrow{AB}\).

\(\overrightarrow{AP} = \frac{ 3 }{ 4 }\left(12\mathbf{b} - 12\mathbf{a}\right) = 9\mathbf{b} - 9\mathbf{a}\)

Now get to \(P\) by going out to \(A\) first:

\(\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = 12\mathbf{a} + 9\mathbf{b} - 9\mathbf{a}\)

\(12\mathbf{a} - 9\mathbf{a} = 3\mathbf{a}\), so

\(\overrightarrow{OP} = 3\mathbf{a} + 9\mathbf{b}\)

Notice the swap: \(AP : PB = 3 : 1\), but the two coefficients come out in the ratio \(1 : 3\) — the other way round. Writing them in the order the ratio is given is the commonest mistake on this question.

Neither \(BC\) nor \(CD\) is printed as an arrow, so build each one through \(A\).

\(\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC}\), and \(\overrightarrow{BA}\) is \(\overrightarrow{AB}\) backwards:

\(\overrightarrow{BC} = -2\mathbf{a} + 2\mathbf{b}\)

\(\overrightarrow{CD} = \overrightarrow{CA} + \overrightarrow{AD} = -2\mathbf{b} + \left(-4\mathbf{a} + 6\mathbf{b}\right)\)

\(\overrightarrow{CD} = -4\mathbf{a} + 4\mathbf{b}\)

\(BCD\) is a straight line only if \(\overrightarrow{CD}\) is a multiple of \(\overrightarrow{BC}\), so match the coefficients.

For \(\mathbf{a}\): \(-2\) becomes \(-4\), a multiplier of \(2\).

For \(\mathbf{b}\): \(2\) becomes \(4\), a multiplier of \(2\).

The multipliers are the same, so CD is a multiple of BC. The two vectors are parallel and share the point C, so BCD is a straight line.

The picture looks straight whichever way the answer goes, so it proves nothing. A bare "Yes" scores nothing; the marks are for the two vectors and for the comparison.

\(OX : XA = 1 : 1\) cuts \(OA\) into \(2\) equal parts with \(OX\) one of them, so \(\overrightarrow{OX} = \frac{1}{ 2 } \times 4\mathbf{a} = 2\mathbf{a}\). In the same way \(\overrightarrow{OY} = \frac{1}{ 2 } \times 2\mathbf{b} = \mathbf{b}\).

Travel from \(X\) to \(Y\) through \(O\):

\(\overrightarrow{XY} = \overrightarrow{XO} + \overrightarrow{OY} = -2\mathbf{a} + \mathbf{b} = \mathbf{b} - 2\mathbf{a}\)

Work out \(\overrightarrow{AB}\) the same way, through \(O\):

\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -4\mathbf{a} + 2\mathbf{b} = 2\mathbf{b} - 4\mathbf{a}\)

Take the \(2\) out as a factor so the two vectors can be compared:

\(\overrightarrow{AB} = 2\left(\mathbf{b} - 2\mathbf{a}\right)\)

Part (a) gave \(\overrightarrow{XY} = \mathbf{b} - 2\mathbf{a}\), so \(\overrightarrow{AB} = 2\,\overrightarrow{XY}\).

One vector being a number times the other is exactly what parallel means, so \(XY\) is parallel to \(AB\).

\(AB\) is \(2\) times as long as \(XY\), so \(XY : AB = 1 : 2\).

Saying "they look parallel" earns nothing. The mark is for writing one vector as a multiple of the other.

Travel from \(A\) to \(C\) through \(O\):

\(\overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = -8\mathbf{a} + 4\mathbf{b} = 4\mathbf{b} - 8\mathbf{a}\)

\(X\) is halfway along \(AC\), so \(\overrightarrow{AX}\) is half of \(\overrightarrow{AC}\):

\(\overrightarrow{AX} = \frac{1}{2}\left(4\mathbf{b} - 8\mathbf{a}\right) = 2\mathbf{b} - 4\mathbf{a}\)

\(\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = 8\mathbf{a} + 2\mathbf{b} - 4\mathbf{a} = 4\mathbf{a} + 2\mathbf{b}\)

First get \(\overrightarrow{OB}\). Go the long way round, through \(A\):

\(AB\) is the side opposite \(OC\), and opposite sides of a parallelogram are equal and parallel, so \(\overrightarrow{AB} = \overrightarrow{OC} = 4\mathbf{b}\).

\(\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = 8\mathbf{a} + 4\mathbf{b}\)

Part (b) gave \(\overrightarrow{OX} = 4\mathbf{a} + 2\mathbf{b}\), and

\(2\,\overrightarrow{OX} = 8\mathbf{a} + 4\mathbf{b} = \overrightarrow{OB}\)

\(\overrightarrow{OB}\) is a multiple of \(\overrightarrow{OX}\), so the two are parallel — and they both start at \(O\), so they are not merely parallel but along the same line. Hence \(O\), \(X\) and \(B\) lie on one straight line.

The multiple is \(2\), so \(X\) is the midpoint of \(OB\) as well as of \(AC\). That is the proof that the diagonals of a parallelogram bisect each other.

A position vector says where a point is; the vector between two points is the difference of their position vectors.

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \left(5\mathbf{a} + 12\mathbf{b}\right) - \left(2\mathbf{a} + 3\mathbf{b}\right) = 3\mathbf{a} + 9\mathbf{b}\)

\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \left(9\mathbf{a} + 24\mathbf{b}\right) - \left(5\mathbf{a} + 12\mathbf{b}\right) = 4\mathbf{a} + 12\mathbf{b}\)

Now compare them:

\(\overrightarrow{BC} = \frac{4}{3}\,\overrightarrow{AB}\)

One is a multiple of the other, so \(AB\) is parallel to \(BC\). Two parallel lines through the same point \(B\) are the same line, so \(A\), \(B\) and \(C\) lie on one straight line.

Both halves are needed for the marks: the multiple gives "parallel", and the shared point \(B\) turns "parallel" into "collinear". Two vectors can be parallel and nowhere near each other.

Get to \(B\) from \(O\) first:

\(\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = 6\mathbf{a} + 4\mathbf{b}\)

Now travel from \(B\) to \(C\) through \(O\):

\(\overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC} = -6\mathbf{a} - 4\mathbf{b} + 12\mathbf{b}\)

\(-4\mathbf{b} + 12\mathbf{b} = 8\mathbf{b}\), so

\(\overrightarrow{BC} = 8\mathbf{b} - 6\mathbf{a}\)

\(M\) is halfway along \(BC\), so \(\overrightarrow{BM}\) is half of \(\overrightarrow{BC}\):

\(\overrightarrow{BM} = \frac{1}{2}\left(8\mathbf{b} - 6\mathbf{a}\right) = 4\mathbf{b} - 3\mathbf{a}\)

Now go out to \(B\) and on to \(M\):

\(\overrightarrow{OM} = \overrightarrow{OB} + \overrightarrow{BM} = 6\mathbf{a} + 4\mathbf{b} + 4\mathbf{b} - 3\mathbf{a}\)

\(\overrightarrow{OM} = 3\mathbf{a} + 8\mathbf{b}\)

There is a quicker route worth knowing, and it makes a good check. \(M\) is the midpoint of \(BC\), so \(\overrightarrow{OM}\) is the average of the two position vectors:

\(\overrightarrow{OM} = \frac{1}{2}\left(\overrightarrow{OB} + \overrightarrow{OC}\right) = \frac{1}{2}\left(6\mathbf{a} + 4\mathbf{b} + 12\mathbf{b}\right) = 3\mathbf{a} + 8\mathbf{b}\)

\(OC\) is \(3\) times as long as \(AB\) and parallel to it, which is what makes \(OABC\) a trapezium rather than a parallelogram.

Travel from \(A\) to \(C\) through \(O\):

\(\overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = -8\mathbf{a} + 2\mathbf{b} = 2\mathbf{b} - 8\mathbf{a}\)

Opposite sides of a parallelogram are equal and parallel, so \(\overrightarrow{AB} = \overrightarrow{OC} = 2\mathbf{b}\) and \(\overrightarrow{BC} = \overrightarrow{AO} = -8\mathbf{a}\).

\(P\) is halfway along \(AB\) and \(Q\) is halfway along \(BC\), so

\(\overrightarrow{PB} = \frac{1}{2} \times 2\mathbf{b} = \mathbf{b}\) and \(\overrightarrow{BQ} = \frac{1}{2} \times \left(-8\mathbf{a}\right) = -4\mathbf{a}\)

\(\overrightarrow{PQ} = \overrightarrow{PB} + \overrightarrow{BQ} = \mathbf{b} - 4\mathbf{a}\)

Put the two answers side by side:

\(\overrightarrow{PQ} = \mathbf{b} - 4\mathbf{a}\) and \(\overrightarrow{AC} = 2\mathbf{b} - 8\mathbf{a}\)

Every coefficient of \(\overrightarrow{AC}\) is twice the matching one, so

\(\overrightarrow{AC} = 2\,\overrightarrow{PQ}\)

One vector is a multiple of the other, which is what parallel means, so \(PQ\) is parallel to \(AC\), and \(PQ : AC = 1 : 2\).

The answer does not depend on which parallelogram it is: joining the midpoints of two sides that meet always gives half of the diagonal joining their far ends.

\(H\) is the midpoint of \(DA\), \(E\) is the midpoint of \(AB\) and \(G\) is the midpoint of \(CD\), so each whole side is twice the given half:

\(\overrightarrow{AD} = 2\mathbf{a}\), \(\overrightarrow{AB} = 4\mathbf{b}\) and \(\overrightarrow{DC} = 10\mathbf{c}\)

First side. Go from \(E\) to \(H\) through \(A\):

\(\overrightarrow{EH} = \overrightarrow{EA} + \overrightarrow{AH} = -2\mathbf{b} + \mathbf{a} = \mathbf{a} - 2\mathbf{b}\)

Opposite side. \(\overrightarrow{BC}\) is not given, so go from \(B\) round through \(A\) and \(D\):

\(\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AD} + \overrightarrow{DC} = -4\mathbf{b} + 2\mathbf{a} + 10\mathbf{c}\)

\(F\) is halfway along \(BC\), so \(\overrightarrow{FC} = \frac{1}{2}\overrightarrow{BC} = \mathbf{a} - 2\mathbf{b} + 5\mathbf{c}\)

\(G\) is halfway along \(CD\), and \(\overrightarrow{CD} = -10\mathbf{c}\), so \(\overrightarrow{CG} = -5\mathbf{c}\)

\(\overrightarrow{FG} = \overrightarrow{FC} + \overrightarrow{CG} = \mathbf{a} - 2\mathbf{b} + 5\mathbf{c} - 5\mathbf{c} = \mathbf{a} - 2\mathbf{b}\)

Both sides came out the same, so \(\overrightarrow{EH} = \overrightarrow{FG}\). They are equal in length and parallel, so \(EFGH\) is a parallelogram.

Nothing was assumed about \(ABCD\), so this holds for every quadrilateral. The other pair works too: \(\overrightarrow{EF} = \overrightarrow{HG} = \mathbf{a} + 5\mathbf{c}\), with the \(\mathbf{b}\) terms cancelling.

There is no arrow given from \(C\) to \(D\), so travel there through \(A\):

\(\overrightarrow{CD} = \overrightarrow{CA} + \overrightarrow{AD}\)

\(\overrightarrow{CA}\) is \(\overrightarrow{AC}\) backwards, so \(\overrightarrow{CA} = -9\mathbf{a} - 8\mathbf{b}\).

\(\overrightarrow{CD} = -9\mathbf{a} - 8\mathbf{b} + 21\mathbf{a} + 12\mathbf{b}\)

Collect the \(\mathbf{a}\) terms and the \(\mathbf{b}\) terms:

\(\overrightarrow{CD} = 12\mathbf{a} + 4\mathbf{b}\)

\(B\), \(C\) and \(D\) are on a straight line, so \(\overrightarrow{BC}\) is a multiple of \(\overrightarrow{CD}\): their coefficients are in the same ratio.

Travel from \(B\) to \(C\) through \(A\):

\(\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC} = -\left(k\mathbf{a} + 6\mathbf{b}\right) + \left(9\mathbf{a} + 8\mathbf{b}\right)\)

\(\overrightarrow{BC} = \left(9 - k\right)\mathbf{a} + \left(8 - 6\right)\mathbf{b}\)

The coefficient without \(k\) is \(2\), and it lines up with \(4\) in \(\overrightarrow{CD}\). The coefficient with \(k\) is \(9 - k\), and it lines up with \(12\). Equate the ratios:

\(\frac{ 9 - k }{ 2 } = \frac{ 12 }{ 4 }\)

\(9 - k = \frac{ 2 \times 12 }{ 4 } = 6\)

\(k = 9 - 6 = 3\)

There is no arrow straight from \(C\) to \(E\), so go from \(E\) to \(C\) through \(D\) and then reverse it:

\(\overrightarrow{EC} = \overrightarrow{ED} + \overrightarrow{DC} = \mathbf{b} + \left(-26\mathbf{a} + 3\mathbf{b}\right) = -26\mathbf{a} + 4\mathbf{b}\)

\(\overrightarrow{CE} = -\overrightarrow{EC} = 26\mathbf{a} - 4\mathbf{b}\)

\(P\) is on \(CE\) extended, with \(CE : EP = 1 : 3\), so \(\overrightarrow{EP}\) is 3 times \(\overrightarrow{CE}\) and in the same direction:

\(\overrightarrow{EP} = 3\overrightarrow{CE} = 3\left(26\mathbf{a} - 4\mathbf{b}\right) = 78\mathbf{a} - 12\mathbf{b}\)

\(P\) is not joined to \(D\) by a given vector, so go through \(E\):

\(\overrightarrow{DP} = \overrightarrow{DE} + \overrightarrow{EP} = -\mathbf{b} + \left(78\mathbf{a} - 12\mathbf{b}\right) = 78\mathbf{a} - 13\mathbf{b}\)

Now \(\overrightarrow{CF}\), going from \(C\) to \(E\) and on to \(F\), with \(\overrightarrow{EF} = -2\mathbf{a}\):

\(\overrightarrow{CF} = \overrightarrow{CE} + \overrightarrow{EF} = \left(26\mathbf{a} - 4\mathbf{b}\right) - 2\mathbf{a} = 24\mathbf{a} - 4\mathbf{b}\)

Take out the common vector from each:

\(\overrightarrow{CF} = -4\left(-6\mathbf{a} +\mathbf{b}\right)\)

\(\overrightarrow{DP} = -13\left(-6\mathbf{a} +\mathbf{b}\right)\)

Both are multiples of the same vector, so \(CF\) is parallel to \(DP\).