Higher January 2019 Paper 2R Q11
11 The straight line \(\mathbf{L}_1\) has equation \(x + 2y = 4\)
The straight line \(\mathbf{L}_2\) passes through the points \((-1, -7)\) and \((7, 9)\)
Michael says that the lines \(\mathbf{L}_1\) and \(\mathbf{L}_2\) are perpendicular.
Is Michael correct?
You must show clearly how you get your answer.
(3)
| Scheme | Marks |
|---|---|
e.g. \(y = 2 - \dfrac{1}{2}x\) or \(y = 2 - \dfrac{x}{2}\) or \(y = \dfrac{4 - x}{2}\) or gradient of \(\mathrm{L}_1 = -0.5\) oe | M1 |
| e.g. \(\dfrac{9 - -7}{7 - -1}\) (=2) or \(\dfrac{-7 - 9}{-1 - 7}\) (=2) | M1 |
| Yes, with correct gradients shown to make −1 when multiplied | A1 |
| (3) | |
| (3 marks) |
Notes
A1: 2 × −0.5 = −1 and yes
| Scheme | Marks |
|---|---|
e.g. \(y = 2 - \dfrac{1}{2}x\) or \(y = 2 - \dfrac{x}{2}\) or \(y = \dfrac{4 - x}{2}\) or gradient of \(\mathrm{L}_1 = -0.5\) oe | M1 |
| \(-7 = 2(-1) + c\) or \(9 = 2(7) + c\) (\(c = -5\)) | M1 |
| Yes, with correct equation shown to be valid by using the given points | A1 |
Notes
M1: dep on M1 for substituting \((-1, -7)\) or \((7, 9)\) into \(y = 2x + c\) to find value of \(c\)
A1: Uses the other point in \(y = 2x - 5\) to show it is valid and yes