To find a dividing point, turn the ratio into a fraction of the whole segment, then write →OP = →OA + (fraction) × →AB. Two lines of working, no formula to recall.
This lesson builds on writing displacements with base vectors and prepares the ground for collinearity.
How do you turn a ratio into a position vector?
- Add the ratio parts to get the denominator. For 2 : 3 that is 5.
- Write →AP as a fraction of →AB. The first part of the ratio is the numerator: →AP = 2/5 →AB.
- Write →AB as b − a.
- Use →OP = →OA + →AP and simplify.
Worked example
→OA = a and →OB = b. P is on AB with AP : PB = 2 : 3. Find →OP.
Step 1: AB is split into 2 + 3 = 5 parts, so →AP = 2/5 →AB.
Step 2: →AB = b − a, so →AP = 2/5(b − a).
Step 3: →OP = →OA + →AP = a + 2/5(b − a).
Step 4: →OP = a − 2/5a + 2/5b = 3/5 a + 2/5 b.
Sense check: P is nearer A, and the coefficient of a (3/5) is larger than that of b (2/5). Also 3/5 + 2/5 = 1, which happens for every point on the line AB.
The mistake to watch for
A common slip is to match the ratio to the vectors in the same order.
Mistaken answer: →OP = 2/5 a + 3/5 b
The student paired “2” with a and “3” with b, because 2 comes first in the ratio.
The sense check catches it: this answer puts more of b into P, so P would sit nearer B. But AP is the smaller part, so P is nearer A. Correct it by deriving →OP = →OA + →AP instead of pairing numbers by position.
Check yourself
1. →OA = a, →OB = b, and P is on AB with AP : PB = 1 : 3. Find →OP.
Show answer
Whole = 4 parts, →AP = 1/4(b − a). →OP = a + 1/4b − 1/4a.
3/4 a + 1/4 b
2. A is (1, 2) and B is (9, 10). P is on AB with AP : PB = 3 : 1. Find the coordinates of P.
Show answer
→AB = (8, 8). →AP = 3/4 × (8, 8) = (6, 6). →OP = (1, 2) + (6, 6).
P = (7, 8)
3. P is on AB and →OP = 5/8 a + 3/8 b. Find AP : PB.
Show answer
Since →OP = a + k(b − a) has the b coefficient equal to k, →AP = 3/8 →AB. So AP is 3 parts and PB is 8 − 3 = 5 parts.
AP : PB = 3 : 5
Where this leads next
With points placed, you can test whether three of them lie on one line in show points are collinear using a scalar relation. Return to the module overview to see the full route.
If ratio questions are where your marks slip, our teachers can rebuild the method with you in online one-to-one Additional Mathematics tuition.