Higher November 2021 Paper 1 Q20
20 The straight line \(\mathbf{L}\) passes through point \(A\ (-6, 2)\) and point \(B\ (5, 3)\)
The straight line \(\mathbf{M}\) is perpendicular to \(\mathbf{L}\) and passes through the midpoint of \(A\) and \(B\).
The line \(\mathbf{M}\) intersects the line \(x = -1\) at point \(C\).
Calculate the area of triangle \(ABC\).
(7)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{-6 + 5}{2}, \dfrac{2 + 3}{2}\right) = \left(-\dfrac{1}{2}, \dfrac{5}{2}\right)\) oe | M1 |
| \(\dfrac{2 - 3}{-6 - 5}\ \left(= \dfrac{-1}{-11} = \dfrac{1}{11}\right)\) oe | M1 |
| \(\dfrac{1}{11} \times m = -1\) or \((m =)\ -11\) | M1ft |
\(\text{``}\tfrac{5}{2}\text{''} = \text{``}{-11}\text{''}\left(\text{``}{-\tfrac{1}{2}}\text{''}\right) + c\) oe or \(y - \text{``}\tfrac{5}{2}\text{''} = \text{``}{-11}\text{''}\left(x - \text{``}{-\tfrac{1}{2}}\text{''}\right)\) and \((y =)\ \text{``}{-11}\text{''}(-1) - 3\ (= 8)\) or \((y =)\ \text{``}{-11}\text{''}\left(-1 - \text{``}{-\tfrac{1}{2}}\text{''}\right) + \text{``}\tfrac{5}{2}\text{''}\ (= 8)\) | M1 |
(See alt methods) \((\text{Perp} =)\ \sqrt{\left(8 - \dfrac{5}{2}\right)^2 + \left(-1 - -\dfrac{1}{2}\right)^2}\ \left(= \dfrac{\sqrt{122}}{2}\right)\) and \((AB =)\ \sqrt{(3 - 2)^2 + (5 - -6)^2}\ (= \sqrt{122})\) | M1 |
| (Area of triangle =) \(\dfrac{1}{2} \times \sqrt{122} \times \dfrac{\sqrt{122}}{2}\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 30.5 | A1 |
| (7) | |
| (7 marks) |
Notes
M1: for finding the midpoint of AB
M1: for finding the gradient of AB
M1ft: their gradient of AB (indep) for the correct use of \(m_1 \times m_2 = -1\)
M1: for an expression that gives the \(y\) value at \(C\)
M1: for a complete method
A1: oe
Allow answers in the range 30.4 – 30.5
| Scheme | Marks |
|---|---|
| Alt 1 (11 × 6) – (0.5 × 1 × 11) – (0.5 × 5 × 6) – (0.5 × 5 × 6) | M1 |
| (11 × 6) – (0.5 × 1 × 11) – (0.5 × 5 × 6) – (0.5 × 5 × 6) | M1 |
Alt 2 \((AC = BC =)\ \sqrt{5^2 + 6^2}\ (= \sqrt{61})\) | M1 |
| (Area of triangle =) \(\dfrac{1}{2} \times \sqrt{61} \times \sqrt{61}\) | M1 |
Alt 3 \(\sqrt{\left(8 - \dfrac{5}{2}\right)^2 + \left(-1 - -\dfrac{1}{2}\right)^2}\ \left(= \dfrac{\sqrt{122}}{2}\right)\) and \((AM =)\ \sqrt{\left(-6 - -\dfrac{1}{2}\right)^2 + \left(2 - \dfrac{5}{2}\right)^2}\ \left(= \dfrac{\sqrt{122}}{2}\right)\) or \((BM =)\ \sqrt{\left(5 - -\dfrac{1}{2}\right)^2 + \left(3 - \dfrac{5}{2}\right)^2}\ \left(= \dfrac{\sqrt{122}}{2}\right)\) | M1 |
| (Area of triangle =) \(2 \times \dfrac{1}{2} \times \dfrac{\sqrt{122}}{2} \times \dfrac{\sqrt{122}}{2}\) | M1 |
Alt 4 \((AC = BC =)\ \sqrt{5^2 + 6^2}\ (= \sqrt{61})\) and \((AB =)\ \sqrt{(3 - 2)^2 + (5 - -6)^2}\ (= \sqrt{122})\) | M1 |
| \(\sqrt{\left(\dfrac{\sqrt{122} + 2\sqrt{61}}{2}\right)\left(\dfrac{\sqrt{122} + 2\sqrt{61}}{2} - \sqrt{122}\right)\left(\dfrac{\sqrt{122} + 2\sqrt{61}}{2} - \sqrt{61}\right)\left(\dfrac{\sqrt{122} + 2\sqrt{61}}{2} - \sqrt{61}\right)}\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) |
Notes
M1: for any 3 correct triangles
M1: for a complete method
M1: for \(AC\) is perp to \(BC\)
M1: for a complete method
M1: for the height of the triangle, \(AM\) and \(BM\) where \(M\) is the midpoint of \(AB\)
M1: for a complete method
M1: for finding \(AC\), \(BC\) and \(AB\)
M1: for applying Heron’s formula