| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Moderate -0.8 This is a straightforward C1 coordinate geometry question testing standard techniques: finding gradient from line equation, parallel lines (same gradient), perpendicular lines (negative reciprocal gradient), and finding intersection points. All steps are routine applications of basic formulas with no problem-solving insight required, making it easier than average but not trivial due to multiple parts. |
| 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 |
|---|---|---|
| \(2x + 3y = 14 \Rightarrow y = \frac{14-2x}{3}\), gradient \(= -\frac{2}{3}\) | M1 A1 | M1 for rearranging to \(y=mx+c\) form or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| DC parallel to AB so gradient \(= -\frac{2}{3}\); through \(D(3,7)\): \(y - 7 = -\frac{2}{3}(x-3)\), i.e. \(2x + 3y = 27\) | M1 A1 | M1 for using \(D(3,7)\) with gradient \(-\frac{2}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of \(AD = \frac{3}{2}\) (negative reciprocal); through \(D(3,7)\): \(y - 7 = \frac{3}{2}(x-3)\); \(2y - 14 = 3x - 9\); \(3x - 2y + 5 = 0\) | M1 A1 M1 A1 | First M1 for using perpendicular gradient condition; second M1 for forming equation through \(D\) |
| Answer | Marks | Guidance |
|---|---|---|
| Solving \(2x + 3y = 14\) and \(5y - x = 6\) simultaneously; from second: \(x = 5y-6\); sub: \(2(5y-6)+3y=14 \Rightarrow 13y=26 \Rightarrow y=2\); \(x=4\); \(B=(4,2)\) | M1 M1 A1 | M1 for attempt to solve simultaneously; M1 for correct substitution |
# Question 1:
## Part (a)
| $2x + 3y = 14 \Rightarrow y = \frac{14-2x}{3}$, gradient $= -\frac{2}{3}$ | M1 A1 | M1 for rearranging to $y=mx+c$ form or equivalent |
## Part (b)(i)
| DC parallel to AB so gradient $= -\frac{2}{3}$; through $D(3,7)$: $y - 7 = -\frac{2}{3}(x-3)$, i.e. $2x + 3y = 27$ | M1 A1 | M1 for using $D(3,7)$ with gradient $-\frac{2}{3}$ |
## Part (b)(ii)
| Gradient of $AD = \frac{3}{2}$ (negative reciprocal); through $D(3,7)$: $y - 7 = \frac{3}{2}(x-3)$; $2y - 14 = 3x - 9$; $3x - 2y + 5 = 0$ | M1 A1 M1 A1 | First M1 for using perpendicular gradient condition; second M1 for forming equation through $D$ |
## Part (c)
| Solving $2x + 3y = 14$ and $5y - x = 6$ simultaneously; from second: $x = 5y-6$; sub: $2(5y-6)+3y=14 \Rightarrow 13y=26 \Rightarrow y=2$; $x=4$; $B=(4,2)$ | M1 M1 A1 | M1 for attempt to solve simultaneously; M1 for correct substitution |
---1 The trapezium $A B C D$ is shown below.\\
\includegraphics[max width=\textwidth, alt={}, center]{66813123-3876-4484-aad1-4bfc09bb1508-2_298_591_557_737}
The line $A B$ has equation $2 x + 3 y = 14$ and $D C$ is parallel to $A B$.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of $A B$.
\item The point $D$ has coordinates $( 3,7 )$.
\begin{enumerate}[label=(\roman*)]
\item Find an equation of the line $D C$.
\item The angle $B A D$ is a right angle. Find an equation of the line $A D$, giving your answer in the form $m x + n y + p = 0$, where $m , n$ and $p$ are integers.
\end{enumerate}\item The line $B C$ has equation $5 y - x = 6$. Find the coordinates of $B$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2010 Q1 [11]}}