June 2022 Paper 2 Q4
4 Line A has equation \(\quad y + 4x = 6\)
Line B is parallel to line A and passes through the point \((2, 1)\)
The point \((d, 2d)\) lies on line B.
Work out the value of \(d\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(y + 4x = c\) or \(y = -4x + c\) or gradient \(= -4\) | M1 | oe \(c\) can be any value other than 6 may be implied |
| \(1 + 4 \times 2 = c\) or \(1 = (\text{their } -4) \times 2 + c\) or \(c = 9\) | M1 | oe their \(-4\) can only be 4 or \(\dfrac{1}{4}\) implied by a correct equation of B eg \(y - 1 = -4(x - 2)\) or \(y + 4x = 9\) or \(y = -4x + 9\) |
| \(2d + 4d = \text{their } 9\) or \(2d = (\text{their } -4)d + \text{their } 9\) or \(6d = 9\) or \(9 \div 6\) | M1dep | oe substitution of \((d, 2d)\) into their equation of B equation with no algebraic denominator dep on 2nd M1 |
| \(\dfrac{3}{2}\) or \(1\dfrac{1}{2}\) or 1.5 | A1 | oe eg \(\dfrac{9}{6}\) |
| Alternative method 2 | ||
| \(y + 4x = c\) or \(y = -4x + c\) or gradient \(= -4\) | M1 | oe \(c\) can be any value other than 6 may be implied |
| \(\dfrac{2d - 1}{d - 2} = \text{their } -4\) | M1 | oe their \(-4\) can only be 4 or \(\dfrac{1}{4}\) may be implied |
| \(2d - 1 = \text{their } -4(d - 2)\) or \(6d = 9\) or \(9 \div 6\) | M1dep | oe equation with no algebraic denominator dep on 2nd M1 |
| \(\dfrac{3}{2}\) or \(1\dfrac{1}{2}\) or 1.5 | A1 | oe eg \(\dfrac{9}{6}\) |
Additional guidance
| Ignore simplification or conversion if correct answer seen | |
| Condone answer (1.5, 3) oe | |
| gradient \(= -4x\) must be recovered | |
| 3rd M1 Allow \((d, 2d)\) to be \((x, 2x)\) etc | |
| 3rd M1 Do not allow use of \((2d, d)\) to be a misread | |
| A correct equation in \(d\) with no algebraic denominator implies M1M1M1 eg \(2d - 1 = -4(d - 2)\) or \(2d = -4d + 9\) or \(6d = 9\) | M1M1M1 |
| Alt 1 gradient = 4 \(y = 4x - 7\) \(2d = 4d - 7 \qquad d = 3.5\) | M0 M1 M1A0 |
| Alt 1 gradient \(= \dfrac{1}{4}\) \(y = \dfrac{1}{4}x + \dfrac{1}{2}\) \(2d = \dfrac{1}{4}d + \dfrac{1}{2} \qquad d = \dfrac{2}{7}\) | M0 M1 M1A0 |
| gradient \(-4\) followed by correct method using gradient 4 or \(\dfrac{1}{4}\) for 2nd and 3rd marks can score a maximum of M2 eg Alt 1 gradient \(-4 \qquad 1 = 4 \times 2 + c \qquad 2d = 4d - 7\) | M0M1M1 |
| gradient \(-4\) followed by correct method using gradient 4 or \(\dfrac{1}{4}\) for 2nd mark (but not the 3rd mark) can score a maximum of M1 eg Alt 1 gradient \(-4 \qquad y = \dfrac{1}{4}x + \dfrac{1}{2}\) (no further valid work) | M0M1M0 |