June 2023 Paper 1 Q7
7 Determine the exact distance between the two points at which the line through \((4, 5)\) and \((6, -1)\) meets the curve \(y = 2x^2 - 7x + 1\). [7]
| Scheme | Marks | AO |
|---|---|---|
| Line through \((4, 5)\) and \((6, -1)\) has gradient \(\frac{-1-5}{6-4} = -3\) So the equation is \(y = 17 - 3x\) | M1 | 3.1a |
| Points of intersection when \(2x^2 - 7x + 1 = 17 - 3x\) | M1 | 1.1a |
| \(2x^2 - 4x - 16 = 0\) | M1 | 1.1b |
| \(x = -2,\ 4\) | A1 | 1.1b |
| when \(x = 4,\ y = 5\) when \(x = -2,\ y = 23\) | A1 | 1.1b |
| distance between \((4, 5)\) and \((-2, 23)\) \(\sqrt{(-2-4)^2 + (23-5)^2}\) | M1 | 1.1a |
| \(= 6\sqrt{10}\) | A1 | 1.1b |
| [7] |
Notes
M1: Attempt to find equation of the line using correct gradient formula
M1: Eliminating one variable
M1: oe Three term quadratic seen or implied by correct \(x\)-values
A1: cao
A1: Both \(y\)-coordinates seen.
M1: Uses distance formula for their points (not given points)
A1: Must be exact (allow \(\sqrt{360}\) oe)