| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Moderate -0.8 This is a straightforward multi-part coordinate geometry question testing standard procedures: midpoint formula, gradient calculation, perpendicular gradient rule, and substitution into a linear equation. All parts are routine textbook exercises requiring only direct application of formulas with no problem-solving insight needed. Slightly easier than average due to the scaffolded structure and 'show that' format in part (c)(i). |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| 1(a) | \(M(3, 2)\) | B1 B1 |
| 1(b) | Gradient \(AB = \frac{-2-6}{5-1} = \frac{-8}{4} = -2\) | M1 A1 |
| 1(c)(i) | Gradient of perpendicular \(= \frac{1}{2}\) | B1 |
| \(\Rightarrow y - 2 = \frac{1}{2}(x-3)\) | M1 | attempt at perp to \(AB\); ft their \(M\) coords |
| \(\Rightarrow 2y - 4 = x - 3 \Rightarrow x - 2y + 1 = 0\) AG | A1 | CSO Must write down the printed answer |
| 1(c)(ii) | \(k - 2(k+5) + 1 = 0\) or \(\frac{(k+5)-2}{k-3} = \frac{1}{2}\) | M1 |
| \(\Rightarrow k = -9\) | A1 | Condone \(x = -9\) (Full marks for correct answer without working) |
**1(a)** | $M(3, 2)$ | B1 B1 | B1 for each coordinate
**1(b)** | Gradient $AB = \frac{-2-6}{5-1} = \frac{-8}{4} = -2$ | M1 A1 | May use coordinates of $M$ instead of $A$ or $B$ - condone one slip. CSO Answer must be simplified to $-2$
**1(c)(i)** | Gradient of perpendicular $= \frac{1}{2}$ | B1 | ft "their" $-1$/gradient $AB$
| $\Rightarrow y - 2 = \frac{1}{2}(x-3)$ | M1 | attempt at perp to $AB$; ft their $M$ coords
| $\Rightarrow 2y - 4 = x - 3 \Rightarrow x - 2y + 1 = 0$ AG | A1 | CSO Must write down the printed answer
**1(c)(ii)** | $k - 2(k+5) + 1 = 0$ or $\frac{(k+5)-2}{k-3} = \frac{1}{2}$ | M1 | Sub into given line equation or correct expression involving gradients. Condone omission of brackets or use of $x$
| $\Rightarrow k = -9$ | A1 | Condone $x = -9$ (Full marks for correct answer without working)
**Total: 9 marks**
---1 The points $A$ and $B$ have coordinates $( 1,6 )$ and $( 5 , - 2 )$ respectively. The mid-point of $A B$ is $M$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $M$.
\item Find the gradient of $A B$, giving your answer in its simplest form.
\item A straight line passes through $M$ and is perpendicular to $A B$.
\begin{enumerate}[label=(\roman*)]
\item Show that this line has equation $x - 2 y + 1 = 0$.
\item Given that this line passes through the point $( k , k + 5 )$, find the value of the constant $k$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2009 Q1 [9]}}