June 2018 Paper 1 Q4
4. The line \(l_1\) has equation \(4y - 3x = 10\)
The line \(l_2\) passes through the points \((5, -1)\) and \((-1, 8)\).
Determine, giving full reasons for your answer, whether lines \(l_1\) and \(l_2\) are parallel, perpendicular or neither. (4)
| Scheme | Marks | AO |
|---|---|---|
| States gradient of \(4y - 3x = 10\) is \(\dfrac{3}{4}\) oe or rewrites as \(y = \dfrac{3}{4}x + \ldots\) | B1 | 1.1b |
| Attempts to find gradient of line joining \((5, -1)\) and \((-1, 8)\) | M1 | 1.1b |
| \(= \dfrac{-1 - 8}{5 - (-1)} = -\dfrac{3}{2}\) | A1 | 1.1b |
| States neither with suitable reasons | A1 | 2.4 |
| (4) | ||
| (4 marks) |
Notes
B1: States that the gradient of line \(l_1\) is \(\dfrac{3}{4}\) or writes \(l_1\) in the form \(y = \dfrac{3}{4}x + \ldots\)
M1: Attempts to find the gradient of line \(l_2\) using \(\dfrac{\Delta y}{\Delta x}\) Condone one sign error Eg allow \(\dfrac{9}{6}\)
A1: For the gradient of \(l_2 = \dfrac{-1 - 8}{5 - (-1)} = -\dfrac{3}{2}\) or the equation of \(l_2\) \(y = -\dfrac{3}{2}x + \ldots\)
Allow for any equivalent such as \(-\dfrac{9}{6}\) or \(-1.5\)
A1: CSO (on gradients)
Explains that they are neither parallel as the gradients not equal nor perpendicular as \(\dfrac{3}{4} \times -\dfrac{3}{2} \neq -1\) oe
Allow a statement in words "they are not negative reciprocals" for a reason for not perpendicular and "they are not equal" for a reason for not being parallel