| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Intersection of two lines |
| Difficulty | Easy -1.2 This is a straightforward multi-part coordinate geometry question testing basic skills: midpoint formula, distance formula, gradient, and solving simultaneous equations. All parts are routine textbook exercises requiring only direct application of standard formulas with no problem-solving insight needed. The 'show that' parts provide the answers, making it easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Midpoint \(= \left(\frac{6+2}{2}, \frac{5-1}{2}\right) = (4,2)\) | M1 | One coordinate correct unsimplified |
| A1 | Both correct and simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(AB^2 = (6-2)^2 + (5+1)^2 = (16+36) = 52\) | M1 | Pythagoras used (condone one slip) |
| A1 | \(52\) or \(\sqrt{52}\) seen | |
| \(\Rightarrow AB = 2\sqrt{13}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Gradient \(AB = (5 - -1)/(6-2)\) | M1 | Must be \(y\) on top and subtraction \((6-2)\) |
| \(= \frac{6}{4} = \frac{3}{2} = 1.5\) | A1 | Any correct equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y - 5 = m(x-6)\) or \(y+1 = m(x-2)\) | M1 | or \(y = mx + c\) and attempt to find \(c\) |
| \(2y - 10 = 3x - 18\) etc leading to \(3x - 2y = 8\) | A1 | AG (be convinced); \(7x = 28\) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Attempt to eliminate \(x\) or \(y\) | M1 | |
| \(x = 4\) | A1 | |
| \(y = 2\) | A1 | \(C\) is point \((4,2)\) |
## Question 1:
**Part (a)**
| Working | Marks | Guidance |
|---------|-------|----------|
| Midpoint $= \left(\frac{6+2}{2}, \frac{5-1}{2}\right) = (4,2)$ | M1 | One coordinate correct unsimplified |
| | A1 | Both correct and simplified |
**Part (b)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $AB^2 = (6-2)^2 + (5+1)^2 = (16+36) = 52$ | M1 | Pythagoras used (condone one slip) |
| | A1 | $52$ or $\sqrt{52}$ seen |
| $\Rightarrow AB = 2\sqrt{13}$ | A1 | |
**Part (c)(i)**
| Working | Marks | Guidance |
|---------|-------|----------|
| Gradient $AB = (5 - -1)/(6-2)$ | M1 | Must be $y$ on top and subtraction $(6-2)$ |
| $= \frac{6}{4} = \frac{3}{2} = 1.5$ | A1 | Any correct equivalent |
**Part (c)(ii)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $y - 5 = m(x-6)$ or $y+1 = m(x-2)$ | M1 | or $y = mx + c$ and attempt to find $c$ |
| $2y - 10 = 3x - 18$ etc leading to $3x - 2y = 8$ | A1 | **AG** (be convinced); $7x = 28$ etc |
**Part (d)**
| Working | Marks | Guidance |
|---------|-------|----------|
| Attempt to eliminate $x$ or $y$ | M1 | |
| $x = 4$ | A1 | |
| $y = 2$ | A1 | $C$ is point $(4,2)$ |
**Total: 12 marks**
---1 The point $A$ has coordinates $( 6,5 )$ and the point $B$ has coordinates $( 2 , - 1 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the midpoint of $A B$.
\item Show that $A B$ has length $k \sqrt { 13 }$, where $k$ is an integer.
\item \begin{enumerate}[label=(\roman*)]
\item Find the gradient of the line $A B$.
\item Hence, or otherwise, show that the line $A B$ has equation $3 x - 2 y = 8$.
\end{enumerate}\item The line $A B$ intersects the line with equation $2 x + y = 10$ at the point $C$. Find the coordinates of $C$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2005 Q1 [12]}}