June 2018 Paper 2 Q2
2 \(P\,(-3,\ -10)\) and \(Q\,(a,\ b)\) are points on a straight line with gradient 12
Work out one possible pair of integer values for \(a\) and \(b\). [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Any pair of integer values for \(a\) and \(b\) for which \(b = 12a + 26\) | B2 | B1 Correct equation in any form eg \(\dfrac{b - -10}{a - -3} = 12\) or \(b + 10 = 12(a + 3)\) or \(\dfrac{y - -10}{x - -3} = 12\) or \(y + 10 = 12(x + 3)\) or \(b = 12a + c\) and \(c = 26\) or \(y = 12x + c\) and \(c = 26\) or \(-3 + \mathrm{k}\) and \(-10 + 12\mathrm{k}\) where k is a non-zero integer |
Additional guidance
| Examples of B2 responses \(a = -4\) and \(b = -22\) or \(a = -2\) and \(b = 2\) or \(a = -1\) and \(b = 14\) or \(a = 0\) and \(b = 26\) or \(a = 1\) and \(b = 38\) or \(a = 2\) and \(b = 50\) or \(a = 3\) and \(b = 62\) or \(a = 4\) and \(b = 74\) | B2 |
| \(a = -3\) and \(b = -10\) is point \(P\) so will not score B2 (B1 possible) | |
| \(-3 + 1\) and \(-10 + 12\) | B1 |
| \(-3 + 2\) and \(-10 + 24\) | B1 |