| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Midpoint of line segment |
| Difficulty | Moderate -0.3 This is a standard C1 coordinate geometry question testing routine techniques: midpoint formula, gradient calculation, equation of a line, and perpendicular gradients. While it has multiple parts (7 marks total), each step follows textbook procedures with no novel insight required. The 'prove' in part (c) simply means showing two gradients multiply to -1. Slightly easier than average due to straightforward calculations and clear structure. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Mid-point of \(BC = (3, -2)\) | B1, B1 | Either coordinate correct; Both cords correct. Accept \(x = 3, y = -2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\Delta y}{\Delta x} = \frac{3-1}{-2-4} = -\frac{1}{3}\) | M1, A1 | \(\pm\frac{2}{6}\) OE implies M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y - 3 = \) "their grad"\((x + 2)\) or \(y - 1 = \) "their grad"\((x - 4)\) Hence \(x + 3y = 7\) | M1, A1 | Or \(y = mx + c\) and correct attempt to find \(c\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y + 5 = \) "their grad \(AB\)"\((x - 2)\) or \(y + 5 = -\frac{1}{3}(x - 2)\) or \(x + 3y + 13 = 0\) | M1, A1 | Or "their \(x + qy = c\)" and attempt to find \(c\); OE |
| Answer | Marks | Guidance |
|---|---|---|
| Grad \(BC = 3\) (from \(\frac{\Delta y}{\Delta x} = \frac{1+5}{4-2}\) OE) \(m_1m_2 = -1\) stated or grad \(BC = 3\) and grad \(AB = -\frac{1}{3}\) or grad \(BC \times\) grad \(AB = (3 \times -\frac{1}{3})\) | B1, M1, A1 CSO | Or 2 lengths correct: \(AB = \sqrt{40}; BC = \sqrt{40}; AC = \sqrt{80}\); Or attempt at Pythagoras or Cosine Rule; \(AC^2 = AB^2 + BC^2 \Rightarrow \angle ABC = 90°\); Completing proof and statement |
| 11 |
## Part (a)
Mid-point of $BC = (3, -2)$ | B1, B1 | Either coordinate correct; Both cords correct. Accept $x = 3, y = -2$
## Part (b)(i)
$\frac{\Delta y}{\Delta x} = \frac{3-1}{-2-4} = -\frac{1}{3}$ | M1, A1 | $\pm\frac{2}{6}$ OE implies M1
## Part (b)(ii)
$y - 3 = $ "their grad"$(x + 2)$ or $y - 1 = $ "their grad"$(x - 4)$ Hence $x + 3y = 7$ | M1, A1 | Or $y = mx + c$ and correct attempt to find $c$
## Part (b)(iii)
$y + 5 = $ "their grad $AB$"$(x - 2)$ or $y + 5 = -\frac{1}{3}(x - 2)$ or $x + 3y + 13 = 0$ | M1, A1 | Or "their $x + qy = c$" and attempt to find $c$; OE
## Part (c)
Grad $BC = 3$ (from $\frac{\Delta y}{\Delta x} = \frac{1+5}{4-2}$ OE) $m_1m_2 = -1$ stated or grad $BC = 3$ and grad $AB = -\frac{1}{3}$ or grad $BC \times$ grad $AB = (3 \times -\frac{1}{3})$ | B1, M1, A1 CSO | Or 2 lengths correct: $AB = \sqrt{40}; BC = \sqrt{40}; AC = \sqrt{80}$; Or attempt at Pythagoras or Cosine Rule; $AC^2 = AB^2 + BC^2 \Rightarrow \angle ABC = 90°$; Completing proof and statement
| | 11 |
---1 The triangle $A B C$ has vertices $A ( - 2,3 ) , B ( 4,1 )$ and $C ( 2 , - 5 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the mid-point of $B C$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the gradient of $A B$, in its simplest form.
\item Hence find an equation of the line $A B$, giving your answer in the form $x + q y = r$, where $q$ and $r$ are integers.
\item Find an equation of the line passing through $C$ which is parallel to $A B$.
\end{enumerate}\item Prove that angle $A B C$ is a right angle.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2008 Q1 [11]}}