June 2023 Paper 1 Q4
4 \(S\,(7, 2)\) and \(T\,(5, -4)\) are points on a straight line.
(a) Work out the gradient of the line. [2 marks]
(b) Work out the distance between \(S\) and \(T\).
Give your answer in the form \(a\sqrt{b}\) where \(a\) and \(b\) are both integers greater than 1 [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{2 - -4}{7 - 5}\) or \(\dfrac{6}{2}\) or \(\dfrac{-4 - 2}{5 - 7}\) or \(\dfrac{-6}{-2}\) | M1 | oe these calculations may appear in a diagram but there needs to be evidence of subtraction and division taking place to gain M1 if A0 |
| 3 | A1 | 3 only (no \(x\)) |
Additional guidance
| Could use \(y = mx + c\) then simultaneous equations in \(m\) and \(c\). If completed correctly then full marks | M1 A1 |
| If the simultaneous equations method is used incorrectly or only partially | M0 A0 |
| Correct answer from graphical method or without working | M1 A1 |
| (\(y =\)) \(3x\) or \(3x - k\) on answer line | M1 A0 |
| Answer | Mark | Comments |
|---|---|---|
| \((7 - 5)^2 + (2 - -4)^2\) or \(2^2 + 6^2\) or \(4 + 36\) | M1 | may be seen on a diagram ft differences from Q4(a) for this mark only (must be numbers used in Q4a and not equivalents – see Additional Guidance) |
| \(\sqrt{40}\) or \(\sqrt{4} \times \sqrt{10}\) | M1dep | |
| \(2\sqrt{10}\) | A1 | just \(a = 2\) and \(b = 10\) would not be enough for A1 (unless \(2\sqrt{10}\) in working) |
Additional guidance
| Penalise incorrect further working | |
| Example of ft from Q4(a) would be \(m = \dfrac{6}{12}\) so [(ST)\(^2\) =] \(6^2 + 12^2\) | M1 M0 A0 |
| But \(m = \dfrac{6}{12}\) then cancelled and so [(ST)\(^2\) =] \(1^2 + 2^2\) | M0 M0 A0 |