To find where two lines cross, write the position of the crossing point twice, once along each line. Then equate the coefficients of a and b.
This lesson builds on finding a ratio from two expressions for one vector. Both belong to vector dependence and ratio reasoning.
What does the question give me?
In triangle OAB, OA = a and OB = b. M lies on OA with OM = (1/3)a, and N is the midpoint of OB, so ON = (1/2)b. The lines AN and BM meet at P. Find OP.
Two unknowns are needed, one for each line. Let AP = λAN and BP = μBM.
Worked solution
Route 1, along AN. OP = OA + λAN. First AN = ON − OA = (1/2)b − a. So OP = a + λ((1/2)b − a) = (1 − λ)a + (λ/2)b.
Route 2, along BM. OP = OB + μBM. First BM = OM − OB = (1/3)a − b. So OP = b + μ((1/3)a − b) = (μ/3)a + (1 − μ)b.
Equate coefficients, since a and b are not parallel.
- a: 1 − λ = μ/3
- b: λ/2 = 1 − μ
From the second equation, λ = 2 − 2μ. Substitute into the first: 1 − (2 − 2μ) = μ/3, so −1 + 2μ = μ/3, and (5/3)μ = 1, which gives μ = 3/5. Then λ = 2 − 6/5 = 4/5.
Now OP = (μ/3)a + (1 − μ)b = (1/5)a + (2/5)b.
Check the answer on both lines
Use λ = 4/5 in Route 1: (1 − 4/5)a + (2/5)b = (1/5)a + (2/5)b. That matches Route 2.
The ratios follow. AP:PN = λ:(1 − λ) = 4/5 : 1/5, which is 4:1. BP:PM = μ:(1 − μ) = 3/5 : 2/5, which is 3:2.
The mistake that loses the marks
The usual slip is to use one letter for both lines. If both routes use λ, the equations become 1 − λ = λ/3 and λ/2 = 1 − λ, which have no common solution, and the work collapses.
| Step | Wrong | Right |
|---|---|---|
| Unknowns | λ on both lines | λ on AN, μ on BM |
| Equations | 1 − λ = λ/3 and λ/2 = 1 − λ | 1 − λ = μ/3 and λ/2 = 1 − μ |
| Outcome | No single solution | λ = 4/5, μ = 3/5 |
A second slip is an error in a direction vector, such as AN = a − (1/2)b. Always subtract start from end: AN = ON − OA.
Check yourself
In triangle OAB, OA = a and OB = b. M is the midpoint of OA, so OM = (1/2)a. N lies on OB with ON = (1/3)b.
Lines AN and BM meet at P. Find OP.
Answer
Route 1, along AN: AN = (1/3)b − a, so OP = a + λ((1/3)b − a) = (1 − λ)a + (λ/3)b.
Route 2, along BM: BM = (1/2)a − b, so OP = b + μ((1/2)a − b) = (μ/2)a + (1 − μ)b.
Equate: 1 − λ = μ/2 and λ/3 = 1 − μ. From the second, λ = 3 − 3μ. Then 1 − 3 + 3μ = μ/2, so (5/2)μ = 2, and μ = 4/5. Then λ = 3 − 12/5 = 3/5.
OP = (2/5)a + (1/5)b.
Check with Route 1: (1 − 3/5)a + (1/5)b = (2/5)a + (1/5)b. It matches.
What to study next
Points can also sit beyond the line segment. That is the topic of checking the sign of an external division ratio, the last lesson in the set.
For help choosing the two routes in a new diagram, see online one-to-one Additional Mathematics tuition. The word-problem structure worksheet gives you a layout for long questions.