Higher June 2025 Paper 1 Q22
22 Here is a sketch of straight line \(EG\).
\(F\) is a point on \(EG\) such that \(\quad EF = \dfrac{1}{3}EG\)
Work out the coordinates of \(F\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Correct method for finding the difference between the \(x\) or \(y\) coordinates for the points \(E\) and \(G\) (positive or negative differences) | M1 | eg \(6 - (-18)\) or 24 or \(3 - 9\) or \(-6\) implied by gradient \(= -\dfrac{1}{4}\) |
| Correct method for finding the difference between the \(x\) or \(y\) coordinates for the points \(E\) and \(F\) or the points \(F\) and \(G\) (positive or negative differences) | M1dep | eg \(\dfrac{-18 - 6}{3}\) or \(-8\) or \(-16\) or \(\dfrac{9 - 3}{3}\) or 2 or 4 |
| (−10, 7) | A1 | SC1 (−2, 5) or (−12, 7.5) oe |
Additional guidance
| SC1 (−2, 5) is for \(EF = \dfrac{2}{3}EG\) | |
| SC1 (−12, 7.5) is for \(EF = \dfrac{1}{3}FG\) | |
| Working may be seen on the diagram | |
| (−10, …) or (…, 7) | M1M1 |