June 2018 Paper 1 Q2
2 \(P\) is the point \((-12,\ b)\)
\(Q\) is the point \((a,\ 4)\)
\(R\) is the point \((6,\ -2)\)
\(Q\) is the midpoint of \(PR\).
Work out the values of \(a\) and \(b\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\dfrac{6 + (-12)}{2}\) or \(\dfrac{-2 + b}{2}\) | M1 | oe eg \(\ 18 \div 2 = 9\) and \(6 - 9\) eg \(\ 6 \times 2 = 12\) and \(-2 + 12\) These come from distances of 18 and 6, as seen in a diagram and used correctly |
| \(a = -3\) | A1 | |
| \(b = 10\) | A1 | |
| Alternative method 2 | ||
| \(a - (-12) = 6 - a\) or \(4 - b = -2 - 4\) | M1 | oe eg \(\ 4 + 4 - -2\) |
| \(a = -3\) | A1 | |
| \(b = 10\) | A1 | |
| Alternative method 3 | ||
| eg \(\ 6 - (-12) = 2(a - -12)\) eg \(\ -2 - b = 2(-2 - 4)\) | M1 | for using an equation relating the "gap" between the points |
| \(a = -3\) | A1 | |
| \(b = 10\) | A1 | |
Additional guidance
| Either answer correct, but no working, implies the M mark, eg \(a = -3\), \(b = 6\) scores M1 A1 A0 | |
| Correct answer seen with no working scores full marks | |
| \(a = 10\) and \(b = -3\) (correct values but the wrong way round) with no working scores SC1 |