June 2017 Paper 2 Q7

7 The equation of the line through \(B\), \(P\) and \(A\) is \(\quad 4x + 5y = 40\)

\(BP : PA = 2 : 3\)

Line 4x + 5y = 40 crossing the y-axis at B and the x-axis at A, with P on BA joined to the origin O

Not drawn accurately

Work out the area of triangle \(OBP\). [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
(\(x\)-coordinate of \(A\) =) 10 and (\(y\)-coordinate of \(B\) =) 8	B1	May be implied on diagram eg 10 written next to \(A\) and 8 written next to \(B\)
(\(x\)-coordinate of \(P\) =) \(\dfrac{2}{2 + 3} \times\) their 10 or \(\dfrac{2 \times \text{their } 10 + 3 \times 0}{2 + 3}\) or 4	M1	oe their 10 must be their \(x\)-coordinate of \(A\) May be seen on diagram
(area of triangle \(OBP\) =) \(\dfrac{1}{2} \times\) their \(8 \times\) their 4	M1dep	oe their 8 must be their \(y\)-coordinate of \(B\)
16	A1ft	ft B0M2
Alternative method 2
(\(x\)-coordinate of \(A\) =) 10 and (\(y\)-coordinate of \(B\) =) 8	B1	May be implied on diagram eg 10 written next to \(A\) and 8 written next to \(B\)
(area of triangle \(OAB\) =) \(\dfrac{1}{2} \times\) their \(10 \times\) their 8 or 40	M1	oe
(area of triangle \(OBP\) =) \(\dfrac{2}{2 + 3} \times\) their 40	M1dep	oe eg their \(40 - \dfrac{3}{2 + 3} \times\) their 40
16	A1ft	ft B0M2
Alternative method 3
(\(x\)-coordinate of \(A\) =) 10 and (\(y\)-coordinate of \(B\) =) 8	B1	May be implied on diagram eg 10 written next to \(A\) and 8 written next to \(B\)
(area of triangle \(OAB\) =) \(\dfrac{1}{2} \times\) their \(10 \times\) their 8 or 40	M1	oe
(\(y\)-coordinate of \(P\) =) \(\dfrac{3}{2 + 3} \times\) their 8 or 4.8 and (area of triangle \(OPA\) =) \(\dfrac{1}{2} \times\) their \(10 \times\) their 4.8 or 24 and (area of triangle \(OBP\) =) their \(40 -\) their 24	M1dep	oe their 8 must be their \(y\)-coordinate of \(B\) \(y\)-coordinate of \(P\) may be seen on diagram
16	A1ft	ft B0M2
Alternative method 4
(\(x\)-coordinate of \(A\) =) 10 and (\(y\)-coordinate of \(B\) =) 8	B1	May be implied on diagram eg 10 written next to \(A\) and 8 written next to \(B\)
(\(AB\) =) \(\sqrt{\text{their } 10^2 + \text{their } 8^2}\) or \(\sqrt{100 + 64}\) or \(\sqrt{164}\) or \(2\sqrt{41}\) or 12.8(…) and (\(BP\) =) \(\dfrac{2}{2 + 3} \times\) their 12.8(…) or 5.12(…) and (angle \(OBP\) =) \(\tan^{-1} \dfrac{\text{their } 10}{\text{their } 8}\) or 51.3(…)	M1	oe their 10 must be their \(x\)-coordinate of \(A\) their 8 must be their \(y\)-coordinate of \(B\)
(area of triangle \(OBP\) =) \(\dfrac{1}{2} \times\) their \(8 \times\) their 5.12 \(\times \sin\) their 51.3	M1dep	oe their 8 must be their \(y\)-coordinate of \(B\)
16	A1ft	ft B0M2

Additional guidance

\(A = 10\) and \(B = 8\)	B1
\(A\) (8, 0) and \(B\) (0, 10) is B0 but can subsequently score up to M2A1ft (answer 16)
\(A\) (0, 10) and \(B\) (8, 0) is B0 but can score up to M2A1ft if uses \(x\)-coordinate of \(A\) as 10 and \(y\)-coordinate of \(B\) as 8 (answer 16)
\(A\) (0, 8) and \(B\) (10, 0) is B0 but can score up to M2A1ft if uses \(x\)-coordinate of \(A\) as 8 and \(y\)-coordinate of \(B\) as 10 (answer 16)
Area triangle \(OBP\) may be seen as the sum of two right-angled triangles
Area triangle \(OBP\) may be seen as area trapezium \(OBPX\) – area triangle \(OPX\) \(X\) is on the \(x\)-axis with \(PX\) perpendicular to the \(x\)-axis
Allow marks for valid working seen even if not subsequently used
15.9(…) \(\to\) answer 16 Answer 15.9(…)	4 marks B1M2A0