Linear Graphs & Gradients: WJEC GCSE Higher November 2021 Unit 1 Q5

Higher November 2021 Unit 1 Q5

(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
\(3k = p - 2\) or \(p - 2 = 3k\) or \(-3k = -\,p + 2\)	B1
\(k = \dfrac{p - 2}{3}\) or \(\dfrac{p - 2}{3} = k\) or \(k = \dfrac{-\,p + 2}{-3}\)	B1	F.T. only from \(\pm 3k = \pm\, p \pm 2\), 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 \(k = (p - 2) \div 3\) 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 \(\frac{3\,+\,7}{2}\) or \(\frac{7\,-\,3}{2} + 3\) OR \(\frac{15\,+\,19}{2}\) or \(\frac{19\,-\,15}{2} + 15\) 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.