Higher June 2025 Paper 2R Q24
24 \(PQ\) is a straight line drawn on a square grid, with a scale of 1 cm for 1 unit on each axis.
\(P\) has coordinates \((-5, a)\) and \(Q\) has coordinates \((7, 3a)\) where \(a \gt 0\)
The length of \(PQ\) is \(4\sqrt{10}\) cm
Find an equation of the perpendicular bisector of \(PQ\)
Give your answer in the form \(y = mx + c\) where \(m\) and \(c\) are integers.
(6)
| Scheme | Marks |
|---|---|
eg \(\sqrt{(3a - a)^2 + (7 - -5)^2} = 4\sqrt{10}\) oe or \((2a)^2 + 12^2 = (4\sqrt{10})^2\) oe or \(4a^2 + 144 = 160\) oe | M1 |
| \(a = \sqrt{\dfrac{\text{``}{160}\text{''} - \text{``}{144}\text{''}}{4}}\;\left(= \sqrt{4} = 2\right)\) | M1 |
\((m_{PQ} =)\;\dfrac{3a - a}{7 - -5}\left(= \dfrac{2a}{12} = \dfrac{a}{6}\right)\) oe or \((m_{PQ} =)\;\dfrac{3 \times [2] - [2]}{7 - -5}\left(= \dfrac{4}{12} = \dfrac{1}{3}\right)\) oe | M1ft |
\(\left[\dfrac{a}{6}\right] \times m_{perp} = -1\) or \((m_{perp} =)\;-1 \div \left[\dfrac{a}{6}\right]\left(= -\dfrac{6}{a}\right)\) oe or \(\left[\dfrac{1}{3}\right] \times m_{perp} = -1\) or \((m_{perp} =)\;-1 \div \left[\dfrac{1}{3}\right]\;(= -3)\) oe | M1ft |
\(\dfrac{-5 + 7}{2}\;(= 1)\) and \(\dfrac{3a + a}{2}\;(= 2a)\) or \(\dfrac{-5 + 7}{2}\;(= 1)\) and \(\dfrac{3 \times [2] + [2]}{2}\;(= 4)\) | M1ft |
| Correct answer only scores full marks (unless from obviously incorrect working) Answer: \(y = -3x + 7\) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: dep on M1
for a complete method to solve a correct equation for \(a\); condone inclusion of \(\pm\)
M1ft: for a method to find the gradient of \(PQ\)
where [2] is what they believe the value of \(a\) to be; must be positive and clearly identified
M1ft: for a method to find the \(x\) coordinate and \(y\) coordinate of the midpoint of \(PQ\); condone if the coordinates are the wrong way around
where [2] is what they believe the value of \(a\) to be; must be positive and clearly identified
A1: oe correct equation in required form eg \(y = 7 - 3x\)