Higher June 2023 Paper 1 Q15
15 The equation of line \(\mathbf{L}_1\) is \(y = 2x - 5\)
The equation of line \(\mathbf{L}_2\) is \(6y + kx - 12 = 0\)
\(\mathbf{L}_1\) is perpendicular to \(\mathbf{L}_2\)
Find the value of \(k\).
You must show all your working. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 3 | P1 | for process to make the \(y\) term the subject of \(6y + kx - 12 = 0\) eg \(6y = 12 - kx\) or \(y = 2 - \dfrac{k}{6}x\) |
| P1 | for process to find gradient of line perpendicular to \(\mathbf{L}_1\), eg \(2 \times m = -1\) or \(m = -\dfrac{1}{2}\) or for process to find \(k\), eg \(-\dfrac{k}{6} \times 2 = -1\) | |
| A1 | cao |
Additional guidance
A correct answer with no supportive working gets 0 marks