Higher June 2024 Paper 2R Q21
21 \(ABCD\) is a square.
The point \(A\) has coordinates \((-5, 2)\)
The point \(B\) has coordinates \((3, 5)\)
Find an equation of the line that passes through \(B\) and \(C\)
Give your answer in the form \(\;ax + by + c = 0\;\) where \(a\), \(b\) and \(c\) are integers.
(4)
| Scheme | Marks |
|---|---|
eg \(\dfrac{5 - 2}{3 - (-5)}\left(= \dfrac{3}{8} = 0.375\right)\) oe or (\(2 = -5m + c\) and \(5 = 3m + c\) leading to) \((m =)\, \dfrac{3}{8}\,(= 0.375)\) or (\(C\) =) (6, −3) or (0, 13) | M1 |
eg \(\left[\dfrac{3}{8}\right] \times m = -1\) oe or \((m =)\, \text{``}{-\dfrac{8}{3}}\text{''}\) oe or \(\dfrac{5 - (-3)}{3 - 6}\left(= -\dfrac{8}{3} = -2.6(666...)\right)\) or \(\dfrac{5 - 13}{3 - 0}\left(= -\dfrac{8}{3} = -2.6(666...)\right)\) | M1ft |
eg \(5 = \text{``}{-\dfrac{8}{3}}\text{''} \times 3 + c\) or \(c = 13\) or \(y = \text{``}{-\dfrac{8}{3}}\text{''}x + 13\) or \(y - 5 = \text{``}{-\dfrac{8}{3}}\text{''}(x - 3)\) oe or \(y - -3 = \text{``}{-\dfrac{8}{3}}\text{''}(x - 6)\) oe or \(y - 13 = \text{``}{-\dfrac{8}{3}}\text{''}(x - 0)\) oe or \(y - 13 = \text{``}{-\dfrac{8}{3}}\text{''}x\) oe | M1ft |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(8x + 3y - 39 = 0\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for a method to find the gradient of \(AB\) or for finding the possible coordinates of \(C\)
M1ft: (indep) for finding the gradient of \(BC\)
Allow perpendicular gradient to be truncated or rounded to 1 dp
\(\left[\dfrac{3}{8}\right]\) means their gradient of \(AB\)
M1ft: (ft dep on previous M1 for their perpendicular gradient)
for substitution to find ‘c’
or
to find an equation for \(BC\)
If students find the coordinates of \(D\) [(–2, –6) or (–8, 10)] then allow for this mark
\(y - -6 = \text{``}{-\dfrac{8}{3}}\text{''}(x - -2)\) oe or
\(y - 10 = \text{``}{-\dfrac{8}{3}}\text{''}(x - -8)\) oe
A1: oe \(a\), \(b\) and \(c\) must be integers
eg. \(16x + 6y = 78\) or \(-8x - 3y + 39 = 0\)
or \(3y = -8x + 39\)