| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Intersection of two lines |
| Difficulty | Moderate -0.8 This is a straightforward C1 coordinate geometry question testing standard techniques: finding gradient from general form, equation of perpendicular line through a point, and solving simultaneous linear equations. All parts are routine textbook exercises requiring only direct application of formulas with no problem-solving insight needed, 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = -\frac{3}{5}x + \frac{11}{5}\) | M1 | Attempt at \(y = f(x)\); or answer \(= \frac{3}{5}\) or \(-\frac{3}{5}x\) gets M1; answer of \(\frac{3}{5}x\) gets M0 |
| Gradient of \(AB = -\frac{3}{5}\) | A1 | Correct answer scores 2 marks. Condone error in rearranging formula if answer for gradient is correct. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(m_1 m_2 = -1\) | M1 | Used or stated |
| Gradient of perpendicular \(= \frac{5}{3}\) | A1\(\checkmark\) | ft their answer from (a)(i) or correct |
| \(y - 1 = \frac{5}{3}(x-2)\) OE | A1 | \(5x - 3y = 7\); or \(y = \frac{5}{3}x + c\), \(c = -\frac{7}{3}\) etc; CSO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Eliminating \(x\) or \(y\), must use \(3x + 5y = 11\) & \(2x + 3y = 8\) | M1 | An equation in \(x\) only or \(y\) only |
| \(x = 7\) | A1 | |
| \(y = -2\) | A1 | Answer only of \((7, -2)\) scores 3 marks |
## Question 1:
### Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = -\frac{3}{5}x + \frac{11}{5}$ | M1 | Attempt at $y = f(x)$; or answer $= \frac{3}{5}$ or $-\frac{3}{5}x$ gets M1; answer of $\frac{3}{5}x$ gets M0 |
| Gradient of $AB = -\frac{3}{5}$ | A1 | Correct answer scores 2 marks. Condone error in rearranging formula if answer for gradient is correct. |
### Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $m_1 m_2 = -1$ | M1 | Used or stated |
| Gradient of perpendicular $= \frac{5}{3}$ | A1$\checkmark$ | ft their answer from (a)(i) or correct |
| $y - 1 = \frac{5}{3}(x-2)$ OE | A1 | $5x - 3y = 7$; or $y = \frac{5}{3}x + c$, $c = -\frac{7}{3}$ etc; CSO |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Eliminating $x$ or $y$, must use $3x + 5y = 11$ & $2x + 3y = 8$ | M1 | An equation in $x$ only or $y$ only |
| $x = 7$ | A1 | |
| $y = -2$ | A1 | Answer only of $(7, -2)$ scores 3 marks |
---1 The line $A B$ has equation $3 x + 5 y = 11$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the gradient of $A B$.
\item The point $A$ has coordinates (2,1). Find an equation of the line which passes through the point $A$ and which is perpendicular to $A B$.
\end{enumerate}\item The line $A B$ intersects the line with equation $2 x + 3 y = 8$ at the point $C$. Find the coordinates of $C$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2009 Q1 [8]}}