Edexcel IGCSE Math A Higher 5 Vectors and Transformation Geometry Questions
Use this Vectors and Transformations hub to move from vector methods into rotations, reflections, translations, enlargements and transformation descriptions.
On the grid, rotate shape P180∘ about the point (-3, 2)
Label the new shape Q.
[ 2 ]
Vertices of the transformed shape: (-9,6), (-9,9), (-3,9), (-6,6)
Question (b)
(b)
On the grid, translate shape P by the vector (810)
Label the new shape R.
Here are triangle T and triangle U drawn on a grid of squares.
[ 1 ]
vertices at (7,3)(10,6)(13,6)(13,3)
Shape in correct position
1
Question (c)
(c)
Describe fully the single transformation that maps triangle T onto triangle U.
[ 3 ]
Transformation: enlargement, scale factor 2, centre (−3,3)
Question 2
[Maximum number: 3]
OAB is a triangle.
OA=4aOB=4bP is the point on AB such that AP:PB=1:3
Question (a)
(a)
Write down AB in terms of a and b
[ 1 ]
Final answer:
AB=−4a+4b
B1, oe, for example 4(b−a).
Question (b)
(b)
Express OP in terms of a and b Give your answer in its simplest form.
[ 2 ]
AB=−4a+4b
OP=4a+41AB
OP=4a+41(−4a+4b)=3a+b
Final answer:
3a+b
M1ft for a correct expression following through their part (a), where AB is of the form ma+nb, m,n=0. A1; allow b+3a.
Question 3
[Maximum number: 3]
Here are two vectors.
FG=(2−5)HG=(144)
Calculate the magnitude of the vector HF
HF=(414)+(5−2)=(912)
∣HF∣=92+122=15
Final answer:
15
M1 for a correct calculation for HF or FH. M1 for a complete magnitude method using their values. A1 from fully correct figures.
Question 4
[Maximum number: 6]
Diagram NOT accurately drawn OABC is a quadrilateral. ABP and ODP are straight lines. OA=2aAB=4b−aOC=5b−3a
(a)
-4 b+a-2 a+5 b-3 a oe
2
M1
for an un-simplified expression for BC in terms of a and b allow a sign error in one term or in one vector eg -4 b+a-2 a-5 b+3 a or -4 b-a-2 a+5 b-3 a or for an answer of 4 a-b
Correct answer scores full marks (unless from obvious incorrect working)
b - 4 a
A1
or -4 a+b
Question (a)
(a)
Find an expression in terms of a and b for the vector BC Simplify your answer.
The point D lies on BC such that B D: D C=1: 3 Given that OP=nOD
[ 2 ]
BC=BA+AO+OC=−(4b−a)−2a+(5b−3a)=b−4a
Final answer: b−4a
M1 for a valid unsimplified expression for BC. A1 oe.
4 ft their BC provided it is in the form m a+n b for all method marks
egOP=2a+4b−a+x(4b−a) or OP=2a+y(4b−a) or OP=5b−3a−( "b −4a " )+λ(4b−a) or OP=5b−3a−43( "b −4a′′)+n("417b′′) or OP="417b′′+n×417b′′ or OP=n×′′417b′′ oe or AP=x(4b−a) or AP=−2a+n×"417b′′ oe or BP=y(4b−a) or BP=41("b−4a′′)+n×"417b′′ oe
M1 for a correct expression, including a parameter, for vector OP, must be clearly identified as OP or for a correct expression, including a parameter, for vector AP,BP,CP or DP must be clearly identified
eg OP=2a+4b−a+x(4b−a) and OP=n×′′417b′′ or OP=2a+y(4b−a) and OP=5b−3a−43( "b −4a " )+n( " 417b " ) or AP=x(4b−a) and AP=−2a+n×′′417b′′ oe or BP=y(4b−a) and BP=41(′′b−4a′′)+n×′′417b′′ for two correct expressions for vector OP or two correct expressions for vector AP, or two correct expressions for vector BP or two correct expressions for vector CP or two correct expressions for vector DP one of which contains n′′417b′′ and one contains x(4 b-a)