4400 Higher November 2008 Paper 4 Q19

EdexcelHigherOld spec4 marksVectors

19 The diagram shows a trapezium \(ABCD\).

Trapezium ABCD with BC parallel to AD; AB marked x and AD marked y

\(\overrightarrow{BC} = 2\overrightarrow{AD}\).
\(\overrightarrow{AB} = \mathbf{x}\). \(\overrightarrow{AD} = \mathbf{y}\).

(a) Find, in terms of \(\mathbf{x}\) and \(\mathbf{y}\),

(i) \(\overrightarrow{AC}\)

(ii) \(\overrightarrow{DC}\)

(2)

(b) The point \(E\) is such that \(\overrightarrow{AE} = \mathbf{x} + \mathbf{y}\).
Use your answer to part (a)(ii) to explain why \(AECD\) is a parallelogram. (2)

Scheme	Marks
(i) Answer: \(\mathbf{x} + 2\mathbf{y}\) oe	B1
(ii) Answer: \(\mathbf{x} + \mathbf{y}\) oe	B1
	(2)

Mark scheme (b)

Scheme	Marks
\(AE = DC\) or \(\overrightarrow{EC} = 2\mathbf{y} - \mathbf{y} = \mathbf{y}\) or \(\overrightarrow{EC} = \mathbf{y}\) shown on diag so \(\overrightarrow{EC} = \overrightarrow{AD}\)	B1
\(AE\) // \(DC\) or \(EC\) & \(AD\) eq & //	B1
	(2)
	(4 marks)

Notes

B1: dep (ii) correct

B1: if no mks otherwise scored, \(\overrightarrow{AE} = \overrightarrow{DC}\) B1