June 2023 Paper 1 Q4

OCR ACurrent spec8 marksVectors

4 It is given that \(ABCD\) is a quadrilateral. The position vector of \(A\) is \(\mathbf{i} + \mathbf{j}\), and the position vector of \(B\) is \(3\mathbf{i} + 5\mathbf{j}\).

(a) Find the length \(AB\). [1]

(b) The position vector of \(C\) is \(p\mathbf{i} + p\mathbf{j}\) where \(p\) is a constant greater than 1.
Given that the length \(AB\) is equal to the length \(BC\), determine the position vector of \(C\). [3]

(c) The point \(M\) is the midpoint of \(AC\).
Given that \(\overrightarrow{MD} = 2\overrightarrow{BM}\), determine the position vector of \(D\). [2]

(d) State the name of the quadrilateral \(ABCD\), giving a reason for your answer. [2]

Mark scheme (a)

Scheme	Marks	AO
\(AB = \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5}\)	B1	1.1
	[1]

Notes

B1: Correct length aef
Condone 4.47 or better
Allow isw eg \(\sqrt{20} = 4\sqrt{5}\)
Allow BOD on signs eg \(AB = -2\mathbf{i} - 4\mathbf{j}\) seen

Mark scheme (b)

Scheme	Marks	AO
\((p - 3)^2 + (p - 5)^2 = 20\)	M1	1.1a
\(p^2 - 8p + 7 = 0\) \(p = 7\)	A1	1.1
\(C\) is \(7\mathbf{i} + 7\mathbf{j}\)	A1	1.1
	[3]

Notes

M1: Attempt correct equation for length \(BC\)
Using their attempt at length of \(AB\)
Condone error on RHS eg having \(\sqrt{20}\) not 20

A1: BC
Solve correct quadratic to obtain at least \(p = 7\)
If second value of \(p\) stated then it must be correct

A1: Correct position vector for \(C\); it could be given as column vector, but not coordinate
No need to discard \(p = 1\)

If M0, question is ‘determine’ so some evidence needed for full marks – either justifying lengths are equal, or use of components of 2 and 4

\(7\mathbf{i} + 7\mathbf{j}\) with some explanation B3
\(7\mathbf{i} + 7\mathbf{j}\) with no explanation B2
(7, 7) with some explanation B2
(7, 7) with no explanation B1

Mark scheme (c)

Scheme	Marks	AO
\(OM\) is \(4\mathbf{i} + 4\mathbf{j}\) OR \(BM\) is \(\mathbf{i} - \mathbf{j}\)	B1	1.1
\(D\) is \(6\mathbf{i} + 2\mathbf{j}\)	B1	1.1
	[2]

Notes

B1: Correct midpoint soi
Could instead find vector \(BM\)
Allow \(M\) seen as coordinate, as it is part of their method and not a requested answer
Condone \(M = 4\mathbf{i} + 4\mathbf{j}\), but penalise clear error eg \(AM = 4\mathbf{i} + 4\mathbf{j}\) is B0
Could be soi on a diagram

B1: Correct position vector (not coordinate) for \(D\)
Do not penalise \(D\) given as coordinate if already penalised in part (b)

Answer only is B0B1

Mark scheme (d)

Scheme	Marks	AO
Kite	B1*	2.2a
eg two pairs of adjacent sides of same length eg diagonals are perpendicular eg \(BD\) being a line of symmetry	B1dep*	2.2a
	[2]

Notes

B1*: Mark independently of reason

B1dep*: Evidence is required to support statements made
All relevant evidence quoted must be correct

Two pairs of adjacent sides: \(AD = CD = \sqrt{26}\) (or compare components of vectors); condone not stating \(AB = BC\) as given in question
Sides must be defined as adjacent, so B0 for just ‘two pairs of equal sides’, but allow BOD if clarified on an explicit diagram seen in part (d)
Diagonals perpendicular: \(AC\) has gradient of 1, \(BD\) has gradient of \(-1\)
If using a geometrical argument, then identify that \(ABC\) is isosceles, \(M\) is mid-point of \(AC\) hence perpendicular bisector
\(BD\) a line of symmetry: \(AM = MC\), with perpendicular argument as above

B0 for reasoning using angles (ie a pair of facing equal angles) unless justified.