Higher June 2025 Paper 1 Q25
25 \(PQRS\) is a square.
\(PR\) is a diagonal of the square.
\(P\) is the point with coordinates \((4, 7)\)
\(R\) is the point with coordinates \((8, -5)\)
Find an equation of the straight line that passes through the points \(Q\) and \(S\)
Give your answer in the form \(ay = bx + c\) where \(a\), \(b\) and \(c\) are integers.
(5)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{4 + 8}{2}, \dfrac{7 - 5}{2}\right)\) oe or (6, 1) | M1 |
| \(\dfrac{7 - -5}{4 - 8}\left(= -\dfrac{12}{4} = -3\right)\) oe | M1 |
| \(\text{``}{-3}\text{''} \times m = -1\) oe or \((m =)\dfrac{-1}{\text{``}{-3}\text{''}}\) or \((m =)\dfrac{1}{3}\) | M1ft |
\(\text{``}{1}\text{''} = \text{``}{\dfrac{1}{3}}\text{''}\left(\text{``}{6}\text{''}\right) + c\) oe or \(c = -1\) or \(y - 1 = \dfrac{1}{3}(x - 6)\) or \(y = \dfrac{1}{3}x - 1\) | M1ft |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \(3y = x - 3\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for finding the midpoint of \(PR\)
M1: for method to find the gradient of \(PR\)
M1ft: for finding the gradient of \(QS\), may be seen embedded in an equation,
ft their gradient of \(PR\)
M1ft: (dep on previous M1)
for finding the equation through \(QS\),
ft their gradient of \(PR\) and their midpoint of \(PR\), do not allow (4, 7) or (8, –5) as midpoint \(PR\)
A1: oe eg \(6y = 2x - 6\) or \(3y - x + 3 = 0\) etc but must be integer coefficients
accept \(a = 3\), \(b = 1\), \(c = -3\)