Higher June 2021 Paper 2 Q17
17 The straight line \(\mathbf{L}\) passes through the points \((4, -1)\) and \((6, 4)\)
The straight line \(\mathbf{M}\) is perpendicular to \(\mathbf{L}\) and intersects the \(y\)-axis at the point \((0, 8)\)
Find the coordinates of the point where \(\mathbf{M}\) intersects the \(x\)-axis.
(4)
| Scheme | Marks |
|---|---|
| eg \(\dfrac{4 - (-1)}{6 - 4}\ \left(= \dfrac{5}{2} = 2.5\right)\) | M1 |
| eg \(\dfrac{-1}{\text{``}{2.5}\text{''}}\ \left(= -\dfrac{2}{5} = -0.4\right)\) or \(\dfrac{-1}{\textit{their}\text{ gradient}}\) oe | M1 |
\(y = \text{``}{−0.4}\text{''}x + 8\) oe eg \(y - 8 = -\dfrac{2}{5}(x - 0)\) or \((8 \div 2) \times 5\ (= 20)\) oe or \(8 \div (-\)‘their gradient of \(\mathbf{M}\)’\()\) | M1 |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: (20, 0) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for a method to find the gradient of \(\mathbf{L}\)
M1: ft for a method to find the gradient of \(\mathbf{M}\) if their gradient of \(\mathbf{L}\) clearly stated (even if no method shown for gradient of \(\mathbf{L}\))
M1: dep on previous M1 for substitution of \((0, 8)\) into equation for a line
or
use of \((8 \div 2) \times 5\ (= 20)\) (maybe on diagram)
NB: 20 gains M3 if clearly intended as \(x\) coordinate (stated or on a diagram)