June 2025 Paper 1 Q4
4 \(AC\) is a straight line.
\(A\) is the point \((-3, 2)\)
\(B\) is the midpoint of the line \(AC\).
\(B\) is the point \((12, 10)\)
(a) Work out the coordinates of \(C\). [2 marks]
(b) Work out the distance \(AB\). [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| (\(x\) coordinate \(=\)) 27 or (\(y\) coordinate \(=\)) 18 or \(12 + (12 - -3)\) or \(10 + (10 - 2)\) or \(-3 + 2(12 - -3)\) or \(2 + 2(10 - 2)\) or \(\begin{pmatrix} 15 \\ 8 \end{pmatrix}\) or \(\begin{pmatrix} 30 \\ 16 \end{pmatrix}\) or \(\dfrac{x - 3}{2} = 12\) or \(\dfrac{y + 2}{2} = 10\) or \(\dfrac{x + 3}{2} = 15\) or \(\dfrac{y - 2}{2} = 8\) | M1 | oe could be seen on a diagram |
| (27, 18) | A1 |
Additional guidance
| 15 and 8 both seen or 30 and 16 both seen (could then be used incorrectly) in working or on diagram | M1 |
| Both coordinates correct but no working | M1A1 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: using AB | ||
| \(\sqrt{\left((12 - -3)^2 + (10 - 2)^2\right)}\) | M1 | oe eg \(\sqrt{289}\) or may spot it’s an 8, 15, 17 triangle |
| 17 | A1 | |
| Alternative method 2: using AC | ||
| \(\sqrt{\left((27 - -3)^2 + (18 - 2)^2\right)}\) | M1 | oe may spot it’s an 8, 15, 17 triangle do not allow ft from incorrect answer to part (a) |
| \((= 34 \div 2 =)\ 17\) | A1 | |
Additional guidance
| Look at part (a) if no marks awarded in part (b) eg \(12 - -3 = 9\) seen in part (a) followed by \(\sqrt{9^2 + 8^2}\) in part (b) | M1A0 |
| B to C would be \(\sqrt{\left((27 - 12)^2 + (18 - 10)^2\right)}\) oe Do not allow ft from incorrect part (a) | M1 |