| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 11 |
| 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 standard techniques: verifying a point lies on a line, finding midpoints, gradients, perpendicular lines, and solving simultaneous linear equations. All parts are routine C1 exercises requiring direct application of formulas with no problem-solving insight needed. Easier than average A-level questions. |
| 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 |
|---|---|---|
| \(21 + 5k = 1 \Rightarrow k = -4\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \((x=) 2\) | B1 | midpoint coords are \((2, -1)\) |
| \((y=) -1\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{1}{5} - \frac{3}{5}x\) | M1 | obtaining \(y = a \pm \frac{3}{5}x\) or \(\frac{\Delta y}{\Delta x} = \frac{-4-2}{7-(-3)} = \frac{-1-2}{2-(-3)} = \frac{-4-(-1)}{7-2}\); condone one sign error in expression |
| (Gradient \(AB =) -\frac{3}{5}\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Perp grad \(= \frac{5}{3}\) | M1 | \(-1/\) "their" grad \(AB\) |
| \(y - 2 = \frac{5}{3}(x + 3)\) or \(y = \frac{5}{3}x + c, c = 7\) etc | A1 | correct equation in any form (must simplify \(x = -3\) to \(x+3\)) or \(c\) to a single term equivalent to 7 |
| \(5x - 3y + 21 = 0\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(3x + 5y = 1\) and \(5x + 8y = 4\) | M1 | must use correct pair of equations and attempt to eliminate \(y\) (or \(x\)) (generous) |
| \(\Rightarrow Px = Q\) or \(Ry = S\) | A1 | |
| \(x = 12, y = -7\) | A1 | 3 marks |
**1(a)(i)**
$21 + 5k = 1 \Rightarrow k = -4$ | B1 | 1 mark | condone $3 \times 7 + 5k = 1$; AG condone $y = -4$
**1(a)(ii)**
$(x=) 2$ | B1 | midpoint coords are $(2, -1)$
$(y=) -1$ | B1 |
**1(b)**
$y = \frac{1}{5} - \frac{3}{5}x$ | M1 | obtaining $y = a \pm \frac{3}{5}x$ or $\frac{\Delta y}{\Delta x} = \frac{-4-2}{7-(-3)} = \frac{-1-2}{2-(-3)} = \frac{-4-(-1)}{7-2}$; condone one sign error in expression
(Gradient $AB =) -\frac{3}{5}$ | A1 | 2 marks | allow $-0.6, \frac{6}{-10}$ etc for A1 & condone error in rearranging if gradient is correct
**1(c)**
Perp grad $= \frac{5}{3}$ | M1 | $-1/$ "their" grad $AB$
$y - 2 = \frac{5}{3}(x + 3)$ or $y = \frac{5}{3}x + c, c = 7$ etc | A1 | correct equation in any form (must simplify $x = -3$ to $x+3$) or $c$ to a single term equivalent to 7
$5x - 3y + 21 = 0$ | A1 | 3 marks | or any multiple of this with integer coefficients –terms in any order but all terms on one side of equation
**1(d)**
$3x + 5y = 1$ and $5x + 8y = 4$ | M1 | must use correct pair of equations and attempt to eliminate $y$ (or $x$) (generous)
$\Rightarrow Px = Q$ or $Ry = S$ | A1 |
$x = 12, y = -7$ | A1 | 3 marks | $(12, -7)$
---1 The point $A$ has coordinates $( - 3,2 )$ and the point $B$ has coordinates $( 7 , k )$.\\
The line $A B$ has equation $3 x + 5 y = 1$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $k = - 4$.
\item Hence find the coordinates of the midpoint of $A B$.
\end{enumerate}\item Find the gradient of $A B$.
\item A line which passes through the point $A$ is perpendicular to the line $A B$. Find an equation of this line, giving your answer in the form $p x + q y + r = 0$, where $p , q$ and $r$ are integers.
\item The line $A B$, with equation $3 x + 5 y = 1$, intersects the line $5 x + 8 y = 4$ at the point $C$. Find the coordinates of $C$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2013 Q1 [11]}}