June 2023 Paper 2 Q10
10 Line K has equation \(\quad 4x - 5y = 17\)
Line L passes through the points \((3, 6)\) and \((-5, 16)\)
Tick (✓) the correct statement about lines K and L.
| The lines are parallel. | |
| The lines are perpendicular. | |
| The lines are neither parallel nor perpendicular. |
Show working to support your answer. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(y = \dfrac{4}{5}x \ldots\) or \(\dfrac{16 - 6}{-5 - 3}\) | M1 | oe eg \(y = \dfrac{-4}{-5}x \ldots\) or \(\dfrac{6 - 16}{3 - -5}\) equation of K must have \(y\) as the subject but the intercept can be ignored allow eg \(y = \dfrac{4x - 17}{5}\) |
| (gradient of K =) \(\dfrac{4}{5}\) or (gradient of L =) \(-\dfrac{5}{4}\) | M1dep | oe value eg (gradient of K =) 0.8 do not allow a gradient if only seen embedded in an equation unless unambiguously selected (eg circled) condone inclusion of \(x\) |
| Both gradients correct and valid explanation and middle box ticked | A1 | eg \(\dfrac{4}{5} \times \dfrac{10}{-8} = -1\) and middle box ticked do not allow a gradient if only seen embedded in an equation unless unambiguously selected (eg circled) do not allow inclusion of \(x\) in either gradient |
Additional guidance
| 0.8 and \(-1.25\) and states gradients multiply to make \(-1\) and middle box ticked | M1M1A1 |
| \(\dfrac{4}{5}\) and \(-\dfrac{5}{4}\) and states negative reciprocals and middle box ticked | M1M1A1 |
| Condone eg \(\dfrac{10}{-8}\) for the value \(-\dfrac{5}{4}\) | |
| M1 may be embedded in equation of line L eg \(y - 6 = \dfrac{16 - 6}{-5 - 3}(x - 3)\) | M1 |
| (gradient of K =) \(\dfrac{4}{5}x \qquad\) (gradient of L =) \(-\dfrac{5}{4}x\) \(\dfrac{4}{5} \times -\dfrac{5}{4} = -1\) and middle box ticked (recovers to use numerical gradients) | M1M1 A1 |
| (gradient of K =) \(0.8x \qquad\) (gradient of L =) \(-1.25\) \(0.8x \times -1.25 = -1\) (includes \(x\)) | M1M1 A0 |