June 2025 Paper 2 Q3
3 The equation of a straight line is \(\quad 2ax + y = 25 \quad\) where \(a\) is a constant.
The line passes through the point \((3, 1)\)
Work out the value of the gradient of the line. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: substitutes (3, 1) into equation of line | ||
| \(6a + 1 = 25\) or \(6a = 24\) | M1 | oe equation eg \(2a \times 3 + 1 = 25\) |
| \((a =)\ 4\) | A1 | implied by \(2a = 8\) or \(-2a = -8\) or \(8x + y = 25\) oe equation of line |
| \(-8\) | B1ft | ft M0A0 or M1A0 ft \(-2 \times\) their \(a\) |
| Alternative method 2: works out the \(y\)-intercept of line | ||
| (0, 25) | M1 | oe eg \(y\)-intercept is 25 |
| \(\dfrac{25 - 1}{0 - 3}\) | M1dep | oe |
| \(-8\) | A1 | |
Additional guidance
| Alt 1 \(\ 2a + 1 = 25 \quad a = 12\) Answer \(-24\) | M0A0 B1ft |
| Alt 1 \(\ 6a = 24 \quad a = 18\) Answer \(-36\) | M1A0 B1ft |
| Answer \(-8x\) or \(-2a\) | B0 |