[Figure 7.1] The diagram shows triangle OAC, where `OA = a`, `OB = b` and `OC = c`. The point B lies on the line AC such that `AB:BC = m:n`, where m and n are constants.
- (a)(i)Write down `AB` in terms of `a` and `b`.[1]
- (a)(ii)Write down `BC` in terms of `b` and `c`.[1]
- (a)(iii)Hence show that `na+mc = (m+n)b`.[2]
- (b)Given that `λ(2/1) + (μ-1)(-4/7) = (λ+1)(4/-2)`, find the value of each of the constants `λ` and `μ`.[4]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.
- [B1] b-a
- [B1] c-b
- [M1] For substitution of their (i) and (ii) into nAB = mBC
- [A1] For correct manipulation to obtain the given answer
- [M1] For equating like components at least once, allow unsimplified
- [M1 Dep] for solving their equations to obtain both λ and μ
- [A1] μ = 5
- [A1] λ = -10
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →