June 2017 Paper 2 Q5
5 \(M\) is the midpoint of \(PQ\).
Not drawn accurately
Work out the value of \(a\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\dfrac{a + 4}{2} = 3a\) or \(3a - a = 4 - 3a\) or \(a + \dfrac{4 - a}{2} = 3a\) or \(4 - \dfrac{4 - a}{2} = 3a\) or \(4 - a = 2(3a - a)\) | M1 | oe |
| \(6a - a = 4\) or \(3a - a + 3a = 4\) or \(2a - a - 6a = -4\) or \(8 - 4 = 6a - a\) or \(4 = 4a + a\) or \(5a = 4\) | M1dep | oe Allow eg \(3a \times 2\) for \(6a\) Terms collected |
| \(\dfrac{4}{5}\) or 0.8 | A1 | oe |
| Alternative method 2 | ||
| \(\dfrac{8 - 6}{3a - a} = \dfrac{10 - 6}{4 - a}\) or \(\dfrac{8 - 6}{3a - a} = \dfrac{10 - 8}{4 - 3a}\) or \(\dfrac{10 - 6}{4 - a} = \dfrac{10 - 8}{4 - 3a}\) | M1 | oe eg fractions inverted |
| \(8a + 2a = 8\) or \(6a + 4a = 8\) or \(-12a + 2a = 8 - 16\) or \(5a = 4\) | M1dep | oe Allow eg \(2a \times 4\) for \(8a\) Terms collected |
| \(\dfrac{4}{5}\) or 0.8 | A1 | oe |
| Alternative method 3 | ||
| \((8 - 6)^2 + (3a - a)^2\) \(= (10 - 8)^2 + (4 - 3a)^2\) or \(5a^2 - 24a + 16 \quad (= 0)\) or \((10 - 6)^2 + (4 - a)^2\) \(= 2^2((8 - 6)^2 + (3a - a)^2)\) or \(15a^2 + 8a - 16 \quad (= 0)\) or \((10 - 6)^2 + (4 - a)^2\) \(= 2^2((10 - 8)^2 + (4 - 3a)^2)\) or \(35a^2 - 88a + 48 \quad (= 0)\) | M1 | oe Using PM\(^2\) = MQ\(^2\) or PQ\(^2\) = 4PM\(^2\) or PQ\(^2\) = 4MQ\(^2\) |
| \((5a - 4)(a - 4) \quad (= 0)\) or \(\dfrac{--24 \pm \sqrt{(-24)^2 - 4 \times 5 \times 16}}{2 \times 5}\) or \((5a - 4)(3a + 4) \quad (= 0)\) or \(\dfrac{-8 \pm \sqrt{8^2 - 4 \times 15 \times -16}}{2 \times 15}\) or \((5a - 4)(7a - 12) \quad (= 0)\) or \(\dfrac{--88 \pm \sqrt{(-88)^2 - 4 \times 35 \times 48}}{2 \times 35}\) | M1dep | oe eg \(\dfrac{12}{5} \pm \sqrt{\dfrac{64}{25}}\) or \(-\dfrac{4}{15} \pm \sqrt{\dfrac{256}{225}}\) or \(\dfrac{44}{35} \pm \sqrt{\dfrac{256}{1225}}\) Correct attempt to solve their 3-term quadratic Allow recovery of brackets in formula Allow eg \(24^2\) for \((-24)^2\) Implied by correct solutions to their 3-term quadratic seen |
| \(\dfrac{4}{5}\) or 0.8 | A1 | oe Must reject solution 4 or \(-\dfrac{4}{3}\) or \(\dfrac{12}{7}\) |
Additional guidance
Terms must be collected but do not have to be processed for M1dep
eg (Alt 1) \(a + 4 = 6a\) needs terms collecting to \(4 = 6a - a\) for M1dep
Rejection of solution is implied by only \(\dfrac{4}{5}\) on answer line