June 2025 Paper 3 Q3
3 The straight line with equation \(2x + y = 6\) crosses the \(x\)-axis at A and the \(y\)-axis at B.
Determine the exact value of \(a\). [3]
| Scheme | Marks | AO |
|---|---|---|
| Straight line drawn | B1 | 1.1 |
| [1] |
Notes
B1: Line through (3, 0) and (0, 6), mark intent
Condone good freehand line
Condone line stopping at axes
Ignore additional lines, perhaps as part of solution to (b) or (c), as long as it is not several attempts at (a)
| Scheme | Marks | AO |
|---|---|---|
| A is \((3, 0)\) and B is \((0, 6)\) | B1 | 1.1 |
| Midpoint is \(\left(\frac{3}{2}, 3\right)\) | M1 | 1.1 |
| Gradient of AB is \(-2\) | M1 | 1.1 |
| Gradient of perpendicular bisector is \(\frac{1}{2}\) | M1 | 1.1 |
| Equation \((y - 3) = \frac{1}{2}\left(x - \frac{3}{2}\right)\) | M1 | 1.1 |
| \(y = \frac{1}{2}x + \frac{9}{4}\) | A1 | 1.1 |
| [6] |
Notes
B1: Coordinates soi by correct midpoint
Condone labels A and B reversed or omitted
Condone missing brackets for coordinates throughout if intent is clear
M1: Correct midpoint or correct method to find the midpoint. e.g. \(\left(\frac{3+0}{2}, \frac{0+6}{2}\right)\)
Allow embedded in the equation for perp. bisector
M1: Correct gradient or correct method to find gradient e.g. \(y = -2x + c\) or \(\frac{6-0}{0-3}\) or \(\frac{6-3}{0-1.5}\) or \(\frac{0-3}{3-1.5}\) oe
FT their midpoint if used
Implied by correct gradient of perpendicular bisector
M1: FT \(\dfrac{-1}{\textit{their}\text{ gradient}}\)
May be seen embedded in the equation.
M1: Using their midpoint and their gradient of perpendicular bisector
Allow \(y = \frac{1}{2}x + c\) and \(3 = \frac{1}{2} \times \frac{3}{2} + c\) oe
A1: OR \(2x - 4y = -9\)
Allow any correct equivalent simplified form e.g. \(4y = 2x + 9\) or \(y - \frac{1}{2}x = \frac{9}{4}\) or \(\frac{1}{2}x - y + \frac{9}{4} = 0\) or any multiple of these.
Allow decimals e.g. \(y = 0.5x + 2.25\)
isw after correct equation seen
| Scheme | Marks | AO |
|---|---|---|
| \((\mathrm{AB}^2 = a^2 + a^2 =)\ 3^2 + 6^2\) Or \((\mathrm{AC}^2 =)\ 1.5^2 + 3^2\) where C is the midpoint of the square | B1 | 3.1a |
| \(a^2 + a^2 = 45\) | M1 | 1.1 |
| \(\frac{3}{2}\sqrt{10}\) oe | A1 | 1.1 |
| [3] |
Notes
B1: Use of Pythagoras which will help to calculate side e.g. AC2, BC2, AB2 or AC, BC, AB soi
FT their midpoint from (a) if used
M1: Further working may not be seen as likely to be done by calculator
A1: Accept any exact correct answer e.g. \(\frac{3\sqrt{10}}{2}\), \(\frac{3}{4}\sqrt{40}\), \(1.5\sqrt{10}\)
Alternative method 1 for M1
| Scheme | Marks |
|---|---|
| \(a^2 = 2\mathrm{AC}^2 = 2(1.5^2 + 3^2)\) | M1 |
M1: Using \(\mathrm{AC} = \sqrt{1.5^2 + 3^2}\) where C is the midpoint of the square
Alternative method 2 for M1
| Scheme | Marks |
|---|---|
| \(a = \sqrt{3^2 + 6^2} \times \cos 45^\circ\) oe | M1 |
M1: Use of trigonometry to find \(a\)
Alternative method 3 for M1
| Scheme | Marks |
|---|---|
| \(a = \dfrac{\sqrt{1.5^2 + 3^2}}{\cos 45^\circ}\) oe | M1 |
M1: Use of trigonometry to find \(a\)
Using \(\mathrm{AC} = \sqrt{1.5^2 + 3^2}\) where C is the midpoint of the square
Alternative method 4 for M1
| Scheme | Marks |
|---|---|
| \((x - 1.5)^2 + (0.5x - 0.75)^2 = 1.5^2 + 3^2\) \(4x^2 - 12x - 27 = 0\) | M1 |
M1: Use of their AC and their equation of perpendicular bisector to form a quadratic equation to give the \(x\)-coordinates of the other two vertices of the square
Alternative method using vectors
| Scheme | Marks |
|---|---|
| \((\overrightarrow{\mathrm{CB}} =) \begin{pmatrix}-1.5\\3\end{pmatrix}\) so \((\overrightarrow{\mathrm{CD}} =) \begin{pmatrix}3\\1.5\end{pmatrix}\) where D is another vertex of the square. | B1 |
| D is \((4.5, 4.5)\) so \(a^2 = (4.5^2 + 1.5^2)\) | M1 |
| \(\frac{3}{2}\sqrt{10}\) oe | A1 |
B1: FT their midpoint from (a)
Vector from C to B or A or vice versa and rotating to get vector to one of the other two vertices
M1: Or \((-1.5, 1.5)\)
A1: Accept any exact correct answer e.g. \(\frac{3\sqrt{10}}{2}\), \(\frac{3}{4}\sqrt{40}\), \(1.5\sqrt{10}\)