Higher June 2017 Paper 6 Q5
5 Point A has coordinates (−4, 6) and point B has coordinates (8, 3).
Not to scale
(a)
(i) Find the gradient of line AB. [2]
(ii) Find the equation of line AB. [2]
(b) Point P has coordinates (0, −2).
Write down the equation of the line parallel to line AB that passes through P. [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) \(-\dfrac{1}{4}\) oe | 2 | M1 for \(\dfrac{\pm(3 - 6)}{\pm(8 - {}^-4)}\) or answer \(\frac{1}{4}\) oe or answer \(-\frac{1}{4}x\) | |
| (ii) \(y = -\frac{1}{4}x + 5\) oe | 2 | M1 for substitution of (−4, 6) or (8, 3) into \(y =\) their (a)(i) \(x + c\) or into \(y - y_1 =\) their (a)(i)\((x - x_1)\) or intercept clearly identified as 5 (may be on diagram or in equation) | eg final answer for 2 marks \(y - 3 = -\frac{1}{4}(x - 8)\) oe or \(y - 6 = -\frac{1}{4}(x - {}^-4)\) oe Missing “\(y =\)” scores M1 max. |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(y = -\frac{1}{4}x - 2\) oe or FT | 2FT | B1FT for \(y =\) their \(mx\) [\(+ a\)] where \(m\) is FT B1 for \(y = bx - 2\), \(b \ne 0\) | FT is for their gradient in (a)(ii) (if no answer in (a)(ii) then use (a)(i)) Condone missing “\(y =\)” if already penalised in (a)(ii), otherwise missing “\(y =\)” is B1 max |