Higher November 2024 Paper 2 Q18
18 The straight line L is perpendicular to the straight line with equation \(2x + y = 9\) and passes through the point with coordinates (8, 11)
Find an equation for L
Give your answer in the form \(y = mx + c\)
(4)
| Scheme | Marks |
|---|---|
| \(y = -2x\;(+9)\) or gradient of line = –2 | M1 |
| Gradient of perpendicular = \(\dfrac{1}{2}\) oe eg \(\dfrac{-1}{-2}\) | M1ft |
| eg \(11 = \text{``}{\tfrac{1}{2}}\text{''} \times 8 + c\) or \(y - 11 = \text{``}{\tfrac{1}{2}}\text{''}(x - 8)\) oe or \(c = 7\) | M1dep |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(y = \dfrac{1}{2}x + 7\) | A1oe |
| (4) | |
| (4 marks) |
Notes
M1: a rearrangement with the correct \(x\) term or stating that the gradient of given line is –2
M1ft: For a statement that the gradient of the perpendicular line is \(\dfrac{1}{2}\) or implication by equation of line with gradient \(\dfrac{1}{2}\)
(if a student goes straight to this stage then M2 is awarded)
This mark can also be awarded for the perpendicular gradient of what they indicate the gradient of the original line to be
M1dep: dep on previous M1 being awarded;
a correct method to find the equation of the perpendicular line by using their gradient of the perpendicular line and (8, 11)
A1oe: a correct equation for the line in the form \(y = mx + c\) (as requested)
If no other marks awarded then award SCB1 for \(y = -\dfrac{1}{2}x + 15\)