O-LevelMathematicsVectors and transformations4024/21 · Oct/Nov 2016 · Q1112 Marks

(a) In the diagram, AB = (-6, 11), AC = (12, -5). [Figure 11.1] (b) [Figure 11.2 - a graph with Triangle A and Triangle B] Triangle A has vertices (½, 1), (1, 2) and (2, 2). Triangle B has vertices (-½, 1), (-1, 2) and (-2, 2).

(a)(i)Find |AC|.[2]
(a)(ii)(a)D is the point such that AD = (0, k), where k > 0. BD is parallel to AC. Show that BD = (6, k-11).[1]
(a)(ii)(b)Find k.[2]
(a)(ii)(c)Find the difference between the lengths of AD and AC.[1]
(b)(i)Describe fully the single transformation that maps triangle A onto triangle B.[2]
(b)(ii)(a)Triangle A is mapped onto triangle C by a transformation represented by the matrix (1 3 / 0 1). Calculate the coordinates of the vertices of triangle C.[2]
(b)(ii)(b)Find the matrix which represents the transformation that maps triangle B onto triangle C.[2]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2016). The mark or mark type is shown in brackets.

[2] (a)(i) 13 M1 for √((-5)² + 12²)
[1] (a)(ii)(a) [BD =] BA + AD = (6, -11) + (0, k) = (6, k-11) AG Or [BD =] AD – AB = (0, k) – (-6, 11) = (6, k-11)
[2] (a)(ii)(b) 8.5 M1 for using 2 x (6 / (k-11)) = (12 / -5)
[1] (a)(ii)(c) 4.5 or FT their (i) – their k
[2] (b)(i) Reflection B1 x = 0 or y-axis B1
[2] (b)(ii)(a) (3½, 1 ), ( 7, 2 ), ( 8, 2) B1 for 1 or 2 correct pairs of coordinates
[2] (b)(ii)(b) (-1 3 / 0 1) B1 for (-1 0 / 0 1) used or M1 for (a b / c d) x ((-1/2) -1 -2 / 1 2 2) = ((3½) 7 8 / 1 2 2)

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