Higher November 2019 Paper 1 Q12
12.
The diagram shows the graph of a straight line, AB.
(a) Find the equation of this line.
Give your answer in the form \(y = mx + c\). [3]
Give your answer in the form \(y = mx + c\). [3]
(b) Find the equation of the perpendicular bisector of the line AB. [4]
| Scheme | Marks | Comments |
|---|---|---|
| \(m = 2\) si | B1 | could be gradient = 2 |
| \(c = 1\) si | B1 | could be \(y\)-intercept = 1 |
| \(y = 2x + 1\) | B1 | Implies all 3 marks |
| Scheme | Marks | Comments |
|---|---|---|
| \(m = -0.5\) si | B1 | FT ‘their gradient AB’; must be clear indication of being the gradient |
| (midpoint = ) (1, 3) | B1 | |
| \(3 = -0.5(1) + c\) | M1 | FT ‘their \(-0.5\)’ provided \(\ne 2\); Accept other full methods e.g. \(\dfrac{y-3}{x-1} = \text{‘their } {-0.5}\text{’}\) or \(y - 3 = \text{‘their } {-0.5}\text{’}(x-1)\) |
| \(y = -0.5x + 3.5\) ISW | A1 | Accept any correct form e.g. \(\dfrac{y-3}{x-1} = -0.5\) or \(y - 3 = -0.5(x-1)\) |
| (7) |