Solving Geometric Problems
Turn geometry into algebra: prove points collinear, divide a line in a ratio, and find unknowns that make vectors parallel — all from one scalar-multiple idea.
Vectors as a problem-solving tool
The real power of vectors is proving geometric facts with algebra. Three ideas do most of the work: two vectors are parallel when one is a scalar multiple of the other; three points are collinear (on one straight line) when the vectors between them are parallel and share a point; and a point divides a line in a ratio when it sits a fixed fraction of the way along. Combine these with $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ and most problems fall out.
Three points $A$, $B$, $C$ are collinear (lie on the same straight line) if $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are parallel — that is, $\overrightarrow{BC} = k\,\overrightarrow{AB}$ for some scalar $k$. Parallel alone isn't enough for separate lines, but here $\overrightarrow{AB}$ and $\overrightarrow{BC}$ already share the point $B$, so if they're also parallel they must lie on one line. This "parallel + common point" test is the standard way to prove collinearity.
Dividing a line in a ratio
If a point $P$ divides $AB$ in the ratio $\mu : \lambda$ (so $AP : PB = \mu : \lambda$), then $P$ is $\dfrac{\mu}{\mu+\lambda}$ of the way from $A$ to $B$:
Start at $A$, then travel the right fraction of the way along $\overrightarrow{AB}$. The midpoint is just the special case $\mu : \lambda = 1 : 1$, giving the $\tfrac12$ from the last lesson.
Vector-proof questions reward a clear method: (1) write each relevant vector using $\mathbf{b} - \mathbf{a}$ and the given information; (2) express the two vectors you're comparing in terms of the same known vectors; (3) show one is a scalar multiple of the other (for parallel/collinear), or match components (to find an unknown). State your conclusion in words — "therefore parallel", "therefore collinear" — to secure the reasoning marks.
Worked example 1
The conclusion sentence earns marks: it's not enough to show $\overrightarrow{BC} = 2\overrightarrow{AB}$ — you must state why that proves collinearity (parallel and a shared point $B$). The scalar $k = 2$ also tells you $C$ is twice as far from $B$ as $B$ is from $A$, in the same direction.
Worked example 2
Two easy-to-confuse steps: the ratio $2:1$ gives the fraction $\tfrac{2}{3}$ (the $2$ over the total $2+1$), and you add $\overrightarrow{AP}$ to $\mathbf{a}$ — not to the origin. $P$ is found by starting at $A$ and stepping $\tfrac23$ of the way along. A sanity check: $P = (6,7)$ should be nearer $B$ than $A$, since $2:1$ puts it two-thirds along — and it is.
Worked example 3
The whole method rests on there being a single $k$ that works for every component. The known component pair ($12$ and $3$) fixes $k = 4$; that same $k$ must then satisfy the unknown pair, giving $t = 5$. This "find $k$ from the known components, apply it to the unknown" pattern solves every "find the value that makes them parallel" question. You can check: $\begin{pmatrix} 12 \\ 20 \end{pmatrix} = 4\begin{pmatrix} 3 \\ 5 \end{pmatrix}$ ✓.
For ratio problems, $\mu : \lambda$ gives the fraction $\tfrac{\mu}{\mu+\lambda}$ — students often use $\tfrac{\mu}{\lambda}$ by mistake. And remember to add the step to the starting point's position vector, not to zero. For collinearity, don't stop at "parallel" — you must note the common point to conclude the points lie on one line. Each of these costs a mark if skipped.
Check yourself
Points $A(1, 1)$, $B(4, 3)$ and $C(10, 7)$ have those coordinates as position vectors. Show that $A$, $B$, $C$ are collinear, and state the ratio $AB : BC$.
(Hint: find $\overrightarrow{AB}$ and $\overrightarrow{BC}$, show one is a multiple of the other, then compare lengths.)
The scalar $2$ does double duty: it proves the vectors are parallel (hence collinear with the shared point $B$), and it gives the length ratio $AB : BC = 1 : 2$ directly.