| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| 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 multi-part coordinate geometry question testing basic skills: rearranging to find gradient, writing parallel line equations, using midpoint formula, and solving simultaneous equations. All parts are routine C1 techniques with no problem-solving insight required, making it easier than average but not trivial due to the multiple steps involved. |
| 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 relationships1.10e Position vectors: and displacement |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{13}{3} - \frac{7}{3}x\) | M1 | attempt at \(y = a + bx\) or \(\frac{\Delta y}{\Delta x}\) with 2 correct points |
| \((gradient) = -\frac{7}{3}\) | A1 | condone slip in rearranging if gradient is correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(y - 3 = \text{'their grad'}(x - (-1))\) | M1 | or \(7x + 3y = k\) and attempt at \(k\) using \(x = -1\) and \(y = 3\) or \(y = (\text{their } m)x + c\) and attempt at \(c\) using \(x = -1\) and \(y = 3\) |
| \(y - 3 = -\frac{7}{3}(x+1)\) or \(7x + 3y = 2\) or \(y = -\frac{7}{3}x + c, c = \frac{2}{3}\) | A1 also | correct equation in any form and replacing \(-\) with \(+\) sign |
| Answer | Marks | Guidance |
|---|---|---|
| \((4, -5)\) | B1, B1 | \(x = 4, y = -5\) withhold if clearly from incorrect working |
| Answer | Marks | Guidance |
|---|---|---|
| \(7x + 3y = 13\) and \(3x + 2y = 12\) \(\Rightarrow\) equation in \(x\) or \(y\) only | M1 | must use correct pair of equations and attempt to eliminate \(y\) (or \(x\)) |
| \(x = -2\) | A1 | |
| \(y = 9\) | A1 |
**1(a)**
$y = \frac{13}{3} - \frac{7}{3}x$ | M1 | attempt at $y = a + bx$ or $\frac{\Delta y}{\Delta x}$ with 2 correct points
$(gradient) = -\frac{7}{3}$ | A1 | condone slip in rearranging if gradient is correct
**1(b)(i)**
$y - 3 = \text{'their grad'}(x - (-1))$ | M1 | or $7x + 3y = k$ and attempt at $k$ using $x = -1$ and $y = 3$ or $y = (\text{their } m)x + c$ and attempt at $c$ using $x = -1$ and $y = 3$
$y - 3 = -\frac{7}{3}(x+1)$ or $7x + 3y = 2$ or $y = -\frac{7}{3}x + c, c = \frac{2}{3}$ | A1 also | correct equation in any form and replacing $-$ with $+$ sign
**1(b)(ii)**
$(4, -5)$ | B1, B1 | $x = 4, y = -5$ withhold if clearly from incorrect working
**1(c)**
$7x + 3y = 13$ and $3x + 2y = 12$ $\Rightarrow$ equation in $x$ or $y$ only | M1 | must use correct pair of equations and attempt to eliminate $y$ (or $x$)
$x = -2$ | A1 |
$y = 9$ | A1 |
---1 The line $A B$ has equation $7 x + 3 y = 13$.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of $A B$.
\item The point $C$ has coordinates $( - 1,3 )$.
\begin{enumerate}[label=(\roman*)]
\item Find an equation of the line which passes through the point $C$ and which is parallel to $A B$.
\item The point $\left( 1 \frac { 1 } { 2 } , - 1 \right)$ is the mid-point of $A C$. Find the coordinates of the point $A$.
\end{enumerate}\item The line $A B$ intersects the line with equation $3 x + 2 y = 12$ at the point $B$. Find the coordinates of $B$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2011 Q1 [9]}}