| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Coordinates from geometric constraints |
| Difficulty | Moderate -0.8 This is a straightforward C1 coordinate geometry question requiring standard techniques: gradient formula, equation of a line, perpendicular gradients, and substitution. All parts follow routine procedures with no problem-solving insight needed, making it easier than average but not trivial since it requires multiple connected steps. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient \(AB = \frac{-1-5}{6-2}\) or \(\frac{5--1}{2-6}\) | M1 | \(\pm\frac{6}{4}\) implies M1 |
| \(= \frac{-6}{4} = -\frac{3}{2}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y-5 = -\frac{3}{2}(x-2)\) or \(y+1 = -\frac{3}{2}(x-6)\) | M1 | or \(y = -\frac{3}{2}x + c\) and attempt to find \(c\) |
| \(\Rightarrow 3x + 2y = 16\) | A1 | OE; must have integer coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient of perpendicular \(= \frac{2}{3}\) | M1 | or use of \(m_1 m_2 = -1\) |
| \(\Rightarrow y - 5 = \frac{2}{3}(x-2)\) | A1 | \(3y - 2x = 11\) (no misreads permitted) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitute \(x = k\), \(y = 7\) into their (b)(i) | M1 | or grads \(\frac{7-5}{k-2} \times \frac{-3}{2} = -1\) |
| \(\Rightarrow 2 = \frac{2}{3}(k-2) \Rightarrow k = 5\) | A1 | or Pythagoras \((k-2)^2 = (k-6)^2 + 8\) |
## Question 1:
### Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient $AB = \frac{-1-5}{6-2}$ or $\frac{5--1}{2-6}$ | M1 | $\pm\frac{6}{4}$ implies M1 |
| $= \frac{-6}{4} = -\frac{3}{2}$ | A1 | AG |
### Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y-5 = -\frac{3}{2}(x-2)$ or $y+1 = -\frac{3}{2}(x-6)$ | M1 | or $y = -\frac{3}{2}x + c$ and attempt to find $c$ |
| $\Rightarrow 3x + 2y = 16$ | A1 | OE; must have integer coefficients |
### Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of perpendicular $= \frac{2}{3}$ | M1 | or use of $m_1 m_2 = -1$ |
| $\Rightarrow y - 5 = \frac{2}{3}(x-2)$ | A1 | $3y - 2x = 11$ (no misreads permitted) |
### Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitute $x = k$, $y = 7$ into their (b)(i) | M1 | or grads $\frac{7-5}{k-2} \times \frac{-3}{2} = -1$ |
| $\Rightarrow 2 = \frac{2}{3}(k-2) \Rightarrow k = 5$ | A1 | or Pythagoras $(k-2)^2 = (k-6)^2 + 8$ |
---1 The points $A$ and $B$ have coordinates $( 6 , - 1 )$ and $( 2,5 )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the gradient of $A B$ is $- \frac { 3 } { 2 }$.
\item Hence find an equation of the line $A B$, giving your answer in the form $a x + b y = c$, where $a , b$ and $c$ are integers.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find an equation of the line which passes through $B$ and which is perpendicular to the line $A B$.
\item The point $C$ has coordinates ( $k , 7$ ) and angle $A B C$ is a right angle.
Find the value of the constant $k$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2007 Q1 [8]}}