June 2024 Paper 1 Q4
4 The equations of two straight lines are
\[y - 3x = 4 \qquad \text{and} \qquad 6y = 18x - 5\]Show that the lines are parallel. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(y = 3x + 4\) from first line or \(y = 3x - \dfrac{5}{6}\) from second line | M1 | oe where one equation is rearranged correctly to match the other eg \(6y = 18x + 24\) from first line |
| \(y = 3x + 4\) and \(y = 3x - \dfrac{5}{6}\) and a statement that they have the same gradient so are parallel | A1 | allow \(y - 3x = 4\) and \(y - 3x = -\dfrac{5}{6}\) eg same gradient so parallel same gradient but different \(y\) intercept so parallel \(m_1 = m_2\) so parallel |
| Alternative method 2 | ||
| 2 correct points on at least one line found and at least one correct calculation using \(\dfrac{y_2 - y_1}{x_2 - x_1}\) to get \(m = 3\) | M1 | eg (1, 7) & (2, 10) or \(\left(0, -\dfrac{5}{6}\right)\) & \(\left(1, \dfrac{13}{6}\right)\) and \(\dfrac{10 - 7}{2 - 1}\) or \(\dfrac{\frac{13}{6} - -\frac{5}{6}}{1 - 0}\) or \(\dfrac{\text{rise}}{\text{run}} = \dfrac{3}{1}\) |
| \(m = 3\) for both lines calculated using \(\dfrac{y_2 - y_1}{x_2 - x_1}\) and a statement that they have the same gradient so are parallel | A1 | eg. same gradient so parallel same gradient but different \(y\) intercept so parallel \(m_1 = m_2\) so parallel must see two separate calculations |
Additional guidance
| In Alt 1 both equations must be correctly rearranged for A mark | |
| Condone \(3x\) as gradient rather than 3 | |
| Gradient \(= 18\) (must rearrange to get gradient of 3 to gain A mark in Alt 1) | A0 |
| Condone \(m\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) used for gradient | |
| A statement that only mentions gradient and doesn’t say they are parallel | A0 |