Higher June 2018 Paper 1 Q20
20 \(ABC\) is an isosceles triangle such that
\(AB = AC\)
\(A\) has coordinates (4, 37)
\(B\) and \(C\) lie on the line with equation \(3y = 2x + 12\)
Find an equation of the line of symmetry of triangle \(ABC\).
Give your answer in the form \(px + qy = r\) where \(p\), \(q\) and \(r\) are integers.
Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
| \(y = \dfrac{2}{3}x \left(+ \dfrac{12}{3}\right)\) or \(y = \dfrac{2x + 12}{3}\) or gradient = \(\dfrac{2}{3}\) | M1 |
(gradient of perpendicular line =) \(-\dfrac{3}{2}\) oe or \(\dfrac{-1}{\text{``}{\frac{2}{3}}\text{''}}\) oe | M1 |
| \(37 = \text{``}{-\dfrac{3}{2}}\text{''} \times 4 + c\) or \(c = 43\) | M1 |
| \(y = -\dfrac{3}{2}x + 43\) | A1 |
| \(3x + 2y = 86\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: ft from their gradient
M1: (dep on previous M1) and ft from their gradient
A1: correct equation (equation in any form)
A1: for \(3x + 2y = 86\) oe for a simplified equation with integer coefficients e.g. \(3x = 86 - 2y\)
Alternatively for the third and fourth marks:
M1 for \(y - 37 = \text{``}{-\dfrac{3}{2}}\text{''}(x - 4)\)
A1 for \(y - 37 = -\dfrac{3}{2}(x - 4)\)
| Scheme | Marks |
|---|---|
| \(2y = -3x + c\) oe | M2 |
| \(2 \times 37 = -3 \times 4 + c\) | M1 |
| \(3x + 2y = 86\) | A2 |
Notes
A2: for \(3x + 2y = 86\) oe for a simplified equation with integer coefficients e.g. \(3x = 86 - 2y\)