Higher June 2022 Paper 2 Q24
24 A straight line
is perpendicular to the straight line through (2, 8) and (6, 15)
and
passes through (0, 9) and (\(x\), 17)
Work out the value of \(x\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{15 - 8}{6 - 2}\) or \(\dfrac{7}{4}\) | M1 | oe eg \(\dfrac{8 - 15}{2 - 6}\) or 1.75 may be embedded in an attempt at equation of line eg \(y = \dfrac{7}{4}x\ldots\) may be implied |
| \(-1 \div\) their \(\dfrac{7}{4}\) or \(-\dfrac{4}{7}\) or \(\dfrac{17 - 9}{x - 0} \times\) their \(\dfrac{7}{4} = -1\) | M1 | oe allow [−0.57143, −0.57] may be embedded in an attempt at equation of a line eg \(y =\) their \(-\dfrac{4}{7}x \ldots\) |
| \(17 - 9 =\) their \(-\dfrac{4}{7}x\) or \(-4x = 56\) or \(56 \div -4\) | M1dep | oe equation must be of the form \(ax = b\) (\(b\) can be unprocessed) dep on 2nd M1 |
| \(-14\) | A1 |
Additional guidance
| The second mark is not dependent on the first – see examples below | |
| (gradient of line through given points =) \(\dfrac{6 - 2}{15 - 8} = \dfrac{4}{7}\) | M0 |
| (gradient of perpendicular line =) \(-\dfrac{7}{4}\) | M1 |
| \(17 - 9 = -\dfrac{7}{4}x\) | M1 |
| (gradient of line through given points =) \(-\dfrac{7}{4}\) | M0 |
| \(\dfrac{17 - 9}{x} \times -\dfrac{7}{4} = -1\) | M1 |
| \(-56 = -4x\) | M1 |
| (gradient of line through given points =) \(\dfrac{7}{4}\) | M1 |
| (gradient of perpendicular line =) \(\dfrac{4}{7}\) | M0M0 |
| Condone use of letters for gradients eg \(x = 1.75\) | M1 |
| For the first two marks, condone inclusion of \(x\) in their gradients | |
| Answer −14 that comes from rounding or truncating cannot score A1 | |
| eg1 (perp grad =) \(-0.57 \qquad 8 = -0.57x \qquad\) Answer \(-14\) | M3A1 |
| eg2 (perp grad =) \(-0.57 \qquad 8 = -0.57x = -14.03 \qquad\) Answer \(-14\) | M3A0 |