Higher January 2023 Paper 2R Q25
25 \(ABCD\) is a trapezium with \(AB\) parallel to \(DC\)
\(A\) is the point with coordinates \((-4, 6)\)
\(B\) is the point with coordinates \((2, 3)\)
\(D\) is the point with coordinates \((-1, 8)\)
The trapezium has one line of symmetry.
The line of symmetry intersects \(CD\) at the point \(E\)
Work out the coordinates of the point \(E\)
(6)
| Scheme | Marks |
|---|---|
| eg \(\left(\dfrac{-4 + 2}{2}, \dfrac{6 + 3}{2}\right)\) or \((-1, 4.5)\) oe | M1 |
| eg \(\dfrac{6 - 3}{-4 - 2}\left(= \dfrac{3}{-6}\right)\) oe or \(-\dfrac{1}{2}\) oe or −0.5 | M1 |
| eg \(m \times \text{``}{-0.5}\text{''} = -1\) oe or \(m = 2\) | M1 |
eg \(y - 8 = \text{``}{-0.5}\text{''}(x - (-1))\) or \(8 = \text{``}{-0.5}\text{''} \times -1 + c\) or \(\dfrac{y - 8}{x - (-1)} = \text{``}{-0.5}\text{''}\) or \(y - 4.5 = \text{``}{2}\text{''}(x - (-1))\) or \(4.5 = \text{``}{2}\text{''} \times -1 + c\) or \(\dfrac{y - 4.5}{x - (-1)} = \text{``}{2}\text{''}\) | M1 |
| eg \(2x + 6.5 = -0.5x + 7.5\) or \(\dfrac{y - 6.5}{2} = \dfrac{y - 7.5}{-0.5}\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: (0.4, 7.3) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for method to find the midpoint of \(AB\)
M1: for method to find the gradient of \(AB\)
M1: for use of \(m_1m_2 = -1\) to find the gradient of the line of symmetry
M1: for method to find an equation for \(CD\) or the line of symmetry
M1: for a correct linear equation to find the \(x\) or \(y\) coordinate of \(E\)
A1: oe
| Scheme | Marks |
|---|---|
| eg \(\dfrac{6 - 3}{-4 - 2}\left(= \dfrac{3}{-6}\right)\) oe or \(-\dfrac{1}{2}\) oe or −0.5 | M1 |
| eg \(y - 8 = \text{``}{-0.5}\text{''}(x + 1)\) or \(8 = \text{``}{-0.5}\text{''} \times -1 + c\) or \(\dfrac{y - 8}{x - (-1)} = \text{``}{-0.5}\text{''}\) | M1 |
| eg \(\sqrt{(-1 - (-4))^2 + (8 - 6)^2}\;(= \sqrt{13})\) | M1 |
| eg \(\sqrt{(x - 2)^2 + (7.5 - 0.5x - 3)^2} = \text{``}{\sqrt{13}}\text{''}\) or \(\sqrt{(15 - 2y - 2)^2 + (y - 3)^2} = \text{``}{\sqrt{13}}\text{''}\) | M1 |
| (1.8, 6.6) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: (0.4, 7.3) | A1 |
Notes
M1: for method to find the gradient of \(AB\)
M1: for method to find an equation for \(CD\)
M1: for method to find the length of \(AD\) or \(AD^2\)
M1: for setting up an equation for the \(x\) or \(y\) coordinate of \(C\)
M1: for the correct coordinates for \(C\)
A1: oe