June 2019 Paper 1 Q1
1 A straight line passes through the points \((-2,\ 11)\) and \((1,\ 2)\)
Work out the equation of the line.
Give your answer in the form \(\quad y = mx + c\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\dfrac{11-2}{-2-1}\) or \(\dfrac{2-11}{1 - -2}\) or \(-3\) | M1 | oe |
| \(11 = (\text{their } -3)(-2) + c\) or \(2 = (\text{their } -3)(1) + c\) or \(c = 5\) | M1 | do not award if \(-3\) from first M mark becomes 3 in this M mark |
| \(y = -3x + 5\) | A1 | condone \(y = 5 - 3x\) |
| Alternative method 2 | ||
| \(\dfrac{11-2}{-2-1}\) or \(\dfrac{2-11}{1 - -2}\) or \(-3\) | M1 | oe |
| \(y - 11 = (\text{their } -3)(x - -2)\) or \(y - 2 = (\text{their } -3)(x - 1)\) | M1 | do not award if \(-3\) from first M mark becomes 3 in this M mark |
| \(y = -3x + 5\) | A1 | condone \(y = 5 - 3x\) |
| Alternative method 3 | ||
| Setting up two simultaneous equations \(11 = -2m + c\) and \(2 = m + c\) | M1 | oe |
| \(m = -3\) or \(c = 5\) | M1dep | must see correct equations |
| \(y = -3x + 5\) | A1 | condone \(y = 5 - 3x\) |
Additional guidance
| \(m = -3\) and/or \(c = 5\) from a diagram | M1, M1 |
| Second M mark is not dependent in alt 1 and alt 2 | |
| Penalise further incorrect work eg \(\ y + 3x = 5\ \) or \(\ y = 2x\) | M1, M1, A0 |