Higher June 2023 Paper 2 Q16
16 Line A
has equation \(\quad y = ax - 1\)
passes through the point (7, 13)
Line B has equation \(\quad 5y - 3x = 4\)
Show that line A has a greater gradient than line B. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(13 = 7a - 1\) or \((a =)\ 2\) | M1 | oe eg \(\dfrac{13 - -1}{7 - 0}\) may be implied eg \((y =)\ 2x - 1\) |
| \((y =)\ \dfrac{3}{5}x \ldots\) or (gradient B \(=\)) \(\dfrac{3}{5}\) | M1 | oe eg (gradient B =) 0.6 allow \((y =)\ \dfrac{3x + 4}{5}\) |
| gradient A = 2 and gradient B \(= \dfrac{3}{5}\) | A1 | oe eg \(2 \gt \dfrac{3}{5}\) condone \(2x \gt \dfrac{3}{5}x\) |
Additional guidance
| Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Condone incorrect \(y\)-intercept eg \(a = 2 \qquad y = \dfrac{3}{5}x + 4\) gradient A = 2 gradient B \(= \dfrac{3}{5}\) | M1M1 A1 |
| It must be clear that the values 2 and \(\dfrac{3}{5}\) are being used to answer the question to award A1 eg1 gradient A = 2 and gradient B \(= \dfrac{3}{5}\) (no statement needed) eg2 \(a = 2 \qquad y = \dfrac{3}{5}x + \dfrac{4}{5}\) eg3 \(y = 2x - 1\) and \(y = \dfrac{3}{5}x + \dfrac{4}{5}\) 2 is greater than \(\dfrac{3}{5}\) eg4 \(y = 2x - 1\) and \(y = \dfrac{3}{5}x + \dfrac{4}{5}\) gradient of A \(\gt\) gradient of B | M2A1 M2A0 M2A1 M2A0 |
| \(13 = 7x - 1\) or \(x = 2\) must be recovered to award 1st M1 |