Higher June 2023 Paper 1 Q12
12.
(a) Circle the equation of a line parallel to the line \(y = 4x + 5\). [1]
\(y = -\dfrac{1}{4}x + 5\) \(y = 4x - 5\) \(y = -4x + 5\) \(y = \dfrac{1}{4}x - 5\) \(4y = x + 5\)
(b) The gradient of the line which passes through the points \((a, 3)\) and \((2a, 9)\) is \(\dfrac{3}{4}\).
Find the value of \(a\). [3]
Find the value of \(a\). [3]
| Scheme | Marks | Comments |
|---|---|---|
| \(y = 4x - 5\) only indicated | B1 |
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{9-3}{2a-a}\) oe, si | B1 | |
| \(\dfrac{9-3}{2a-a} = \dfrac{3}{4} \times \left(\dfrac{2}{2}\right)\) oe, si | M1 | FT ‘their derived gradient’ providing numerator or denominator correct. \(2a - a = 8\) Award B1 M1 for change in \(y\) = 6 and change in \(x\) = 8. |
| \((a =)\ 8\) | A1 | CAO |
| (4) |