June 2019 Paper 1 Q8
8 \(ABC\) is a straight line.
\(A\) is the point \((-4,\ 5)\)
\(C\) is the point \((20,\ -7)\)
\(AB : BC = 5 : 3\)
Work out the coordinates of \(B\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\pm(20 - -4)\) or \(\pm(5 - -7)\) or \(\pm 24\) or \(\pm 12\) seen | M1 | allow on diagram |
| using \(\dfrac{5}{8}\) or \(\dfrac{3}{8} \times \pm\) their 24 or \(\pm 15\) or \(\pm 9\) or \(\dfrac{5}{8}\) or \(\dfrac{3}{8} \times \pm\) their 12 or \(\pm 7.5\) or \(\pm 4.5\) | M1dep | oe |
| \((11,\ -2.5)\) | A2 | A1 for each |
| Alternative method 2 | ||
| \((x =)\ \dfrac{3(-4) + 5(20)}{8}\) or \((y =)\ \dfrac{3(5) + 5(-7)}{8}\) | M1 | oe (condone 1 numerical error) |
| \((x =)\ \dfrac{3(-4) + 5(20)}{8}\) and \((y =)\ \dfrac{3(5) + 5(-7)}{8}\) | M1 | oe |
| \((11,\ -2.5)\) | A2 | A1 for each |
Additional guidance
| \((6,\ 0.5)\) if no other marks gained (from 3 and 5 reversed) | SC1 |
| 11 or \(-2.5\) seen in answer line with no working | M1, M1, A1, A0 |
| \((-2.5,\ 11)\) without working can be awarded the method marks | M1, M1, A0, A0 |
| 11 or \(-2.5\) coming from correct working can be awarded one A mark but for A2 these need to be written as coordinates in brackets |