Higher November 2018 Paper 4 Q18
18 P is the point (0, –1) and Q is the point (5, 9).
Find the equation of the line through P that is perpendicular to the line PQ. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(y = -\frac{1}{2}x - 1\) oe | 5 | B2 for gradient 2 or M1 for \(\dfrac{\pm(9 - -1)}{\pm(5 - 0)}\) or gradient of –2 AND M1 for ‘\(m\)’ = \(\dfrac{-1}{\textit{their } 2}\) B1 for \(-\frac{1}{2}x - 1\), \(y = -\frac{1}{2}x + c\) or \(y = mx - 1\) or \(y = (\textit{their } m)x + c\) as answer | |