November 2021 Paper 2 Q12
12 \(A\) and \(B\) are points on the line \(\quad y = 3x + 2\)
\(B\), \(C\) and \(D\ (5, 0)\) are points on the line L.
\(OA : AC = 1 : 4\)
Not drawn accurately
Work out the \(x\)-coordinate of \(B\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| 10 | B1 | \(y\)-coordinate of \(C\) may be seen on the graph |
| \((-)\dfrac{\text{their } 10}{5}\) or \((-)\ 2\) | M1 | ± their gradient of L |
| \((y =)\ -\dfrac{\text{their } 10}{5}x + \text{their } 10\) | M1dep | oe eg \(y - 0 = -\dfrac{\text{their } 10}{5}(x - 5)\) or \(y - \text{their } 10 = -\dfrac{\text{their } 10}{5}(x - 0)\) must use a negative gradient |
| \(-\dfrac{\text{their } 10}{5}x + \text{their } 10 = 3x + 2\) or \(5x = 8\) | M1dep | oe |
| 1.6 | A1ft | oe eg \(\dfrac{8}{5}\) ft B0M3 |
Additional guidance
| A1ft values must be exact or rounded to 1 decimal place or better | |
| Ignore any \(y\)-coordinate of \(B\) calculated after working out the \(x\)-coordinate | |
| Assuming the lines are perpendicular can score a maximum of B1 | |
| \(y\)-coordinate of \(C = 8\) gradient L \(= -\dfrac{8}{5}\) \(y = -\dfrac{8}{5}x + 8\) \(-\dfrac{8}{5}x + 8 = 3x + 2\) 1.3 \(\left(\text{Note that the exact value is } \dfrac{30}{23}\right)\) | B0 M1 M1 M1 A1ft |