When P divides AB in the ratio m:n, P is m out of m + n equal steps from A to B. So OP = a + (m/(m + n))(b − a), where a = OA and b = OB.
This lesson is part of SPM Additional Mathematics vectors. You need parallel vectors and collinearity first, because P lying on AB is a collinearity statement.
How do I find OP when AP:PB = 2:3?
Go from O to A, then along AB. The ratio tells you how far along AB to go.
- AP:PB = 2:3 means the line is 2 + 3 = 5 equal parts, and AP takes 2 of them. So AP = (2/5)AB.
- Write the route: OP = OA + AP = a + (2/5)AB.
- Replace AB with b − a: OP = a + (2/5)(b − a) = a − (2/5)a + (2/5)b.
- Collect terms: OP = (3/5)a + (2/5)b.
The coefficients add up to 3/5 + 2/5 = 1, which is the signal that P lies on AB. The larger weight is on a because P is closer to A.
Does it work with real numbers?
Test with a = (5, 10) and b = (15, 0). Then OP = (3/5)(5, 10) + (2/5)(15, 0) = (3, 6) + (6, 0) = (9, 6).
Now check the ratio. AP = (9 − 5, 6 − 10) = (4, −4), and PB = (15 − 9, 0 − 6) = (6, −6). The lengths are in the ratio 4:6, which is 2:3, as required.
The mistake that loses the marks
The usual slip is to use the ratio as a fraction of AB without adding the parts. For AP:PB = 2:3, it gives AP = (2/3)AB.
| Step | Wrong | Right |
|---|---|---|
| Fraction of AB | 2/3 | 2/5 |
| OP | a + (2/3)(b − a) = (1/3)a + (2/3)b | (3/5)a + (2/5)b |
| Test with a = (5, 10), b = (15, 0) | (1/3)(5, 10) + (2/3)(15, 0) = (11 2/3, 3 1/3) | (9, 6) |
The wrong point gives AP = (6 2/3, −6 2/3) and PB = (3 1/3, −3 1/3). Their lengths are in ratio 2:1, not 2:3, so the test exposes the error.
A second slip is to read the ratio backwards. Mark AP and PB on the diagram first, in the order the question writes them.
Check yourself
The point Q lies on AB such that AQ:QB = 1:3. Express OQ in terms of a and b.
Answer
The line is 1 + 3 = 4 equal parts, and AQ takes 1 of them, so AQ = (1/4)AB.
OQ = OA + AQ = a + (1/4)(b − a) = (3/4)a + (1/4)b.
The coefficients add up to 1, and Q is closer to A, which matches a larger weight on a.
What to study next
When two lines cross inside a diagram, you need two such routes to the same point. That is the topic of vector dependence and ratio reasoning, and checking the sign of an external division ratio covers points outside the line segment.
If ratio questions still feel like a separate memorised rule, online one-to-one Additional Mathematics tuition lets a teacher rebuild the derivation on your questions. The word-problem structure worksheet also helps you set out a long question before you start.