Higher June 2018 Paper 1 Q19
19 The point \(P\) has coordinates \((3, 4)\)
The point \(Q\) has coordinates \((a, b)\)
A line perpendicular to \(PQ\) is given by the equation \(3x + 2y = 7\)
Find an expression for \(b\) in terms of \(a\). (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(b = \dfrac{2}{3}a + 2\) | P1 | for process to rearrange the equation to give \(y\) in terms of \(x\) eg \(y = \dfrac{7 - 3x}{2}\) or \(y = -\dfrac{3}{2}x + \left(\dfrac{7}{2}\right)\) or \(m = -\dfrac{3}{2}\) |
| P1 | for using their gradient in \(mn = -1\) | |
| P1 | for showing a process to find the gradient of \(PQ\) eg \(\dfrac{b - 4}{a - 3}\) OR for substituting \(x = 3\) and \(y = 4\) in \(y = \text{``}\tfrac{2}{3}\text{''}x + c\) | |
| P1 | (dep P3) for forming an equation in \(a\) and \(b\) eg \(\dfrac{b - 4}{a - 3} = \text{``}\tfrac{2}{3}\text{''}\) or \(b = \text{``}\tfrac{2}{3}\text{''}a + \text{``}2\text{''}\) OR correct equation in terms of \(x\) and \(y\) eg \(y = \dfrac{2}{3}x + 2\) | |
| A1 | for \(b = \dfrac{2}{3}a + 2\) oe |
Additional guidance
\(y - 4 = \dfrac{2}{3}(x - 3)\) gets P4
Accept 0.66 or 0.67 oe for \(\dfrac{2}{3}\)