Linear Graphs & Gradients: WJEC GCSE Intermediate November 2021 Unit 1 Q14

Intermediate November 2021 Unit 1 Q14

14.

(a) Rearrange the following formula to make \(k\) the subject.\[p = 3k + 2\][2]

(b) Does the midpoint of the straight line joining points (3, 15) and (7, 19) lie on the line \(y = 3x + 2\)?
You must show all your working. [3]

Mark scheme (a)

Scheme	Marks	Comments
3\(k\) = \(p\) − 2 or \(p\) − 2 = 3\(k\) or –3\(k\) = −\(p\) + 2	B1
\(k = \dfrac{p - 2}{3}\) or \(\dfrac{p - 2}{3} = k\) or \(k = \dfrac{-p + 2}{-3}\)	B1	F.T. only from ±3\(k\) = ±p ± 2, stated or implied. (3k = \(p\) − 2 will have already gained the previous B1.) B1B0 for − \(k = \dfrac{-p + 2}{3}\) or equivalent. Mark final answer. Note Allow B1B0 for k = (p − 2) ÷ 3 with or without brackets. Allow B1B0 for \(\dfrac{p - 2}{3}\) (‘k’ missing)

Mark scheme (b)

Scheme	Marks	Comments
(Midpoint =) (5, 17)	B2	B1 for each coordinate. May be given as \(x\) = 5 and \(y\) = 17. Accept use of \(x\) = 5 and \(y\) = 17 in \(y = 3x + 2\). Allow B1 for sight of \(\dfrac{3+7}{2}\) or \(\dfrac{7-3}{2} + 3\) OR \(\dfrac{15+19}{2}\) or \(\dfrac{19-15}{2} + 15\) Allow SC1 for unsupported (17, 5).
Showing that 17 = 3 × 5 + 2 (convincing) AND ‘Yes’	B1	FT ‘their stated midpoint’, but not (3,15) nor (7,19), with consequent calculation AND decision.