| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 10 |
| 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 multi-part coordinate geometry question testing basic skills: verifying a point lies on a line, finding midpoints, gradients, perpendicular gradients, and line 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 | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| \(15 + 4k = 7 \Rightarrow 4k = -8 \Rightarrow k = -2\) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}(x_1 + x_2)\) or \(\frac{1}{2}(y_1 + y_2)\) | M1 | — |
| Midpoint coordinates \(\left(3, -\frac{1}{2}\right)\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt at \(\Delta y/\Delta x\) or \(y = -\frac{3}{4}x + \frac{7}{4}\) | M1 | (Not x over y)(may use M instead of A/B) |
| Gradient \(AB = -\frac{3}{4}\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(m_1 m_2 = -1\) used or stated | — | 1 |
| Hence gradient \(AC = \frac{4}{3}\) | A1√ | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y - 1 = \frac{4}{3}(x-1)\) or \(3y = 4x - 1\) etc | B1√ | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 0 \Rightarrow x - 1 = -\frac{3}{4}\) | M1 | Putting \(y = 0\) in their AC equation and attempting to find x |
| \(x = \frac{1}{4}\) | A1 | 2 |
## 2(a)(i)
$15 + 4k = 7 \Rightarrow 4k = -8 \Rightarrow k = -2$ | B1 | 1 | AG (condone verification or $y = -2$)
## 2(a)(ii)
$\frac{1}{2}(x_1 + x_2)$ or $\frac{1}{2}(y_1 + y_2)$ | M1 | —
Midpoint coordinates $\left(3, -\frac{1}{2}\right)$ | A1 | 2 | One coordinate correct implies M1
## 2(b)
Attempt at $\Delta y/\Delta x$ or $y = -\frac{3}{4}x + \frac{7}{4}$ | M1 | (Not x over y)(may use M instead of A/B)
Gradient $AB = -\frac{3}{4}$ | A1 | 2 | $-0.75$ etc any correct equivalent
## 2(c)(i)
$m_1 m_2 = -1$ used or stated | — | 1 | —
Hence gradient $AC = \frac{4}{3}$ | A1√ | 2 | Follow through their gradient of AB from part (b)
## 2(c)(ii)
$y - 1 = \frac{4}{3}(x-1)$ or $3y = 4x - 1$ etc | B1√ | 1 | Follow through their gradient of AC from part (c)(i) must be **normal** & (1,1) used
## 2(c)(iii)
$y = 0 \Rightarrow x - 1 = -\frac{3}{4}$ | M1 | Putting $y = 0$ in their AC equation and attempting to find x
$x = \frac{1}{4}$ | A1 | 2 | CSO. C has coordinates $\left(\frac{1}{4}, 0\right)$
---2 The point $A$ has coordinates $( 1,1 )$ and the point $B$ has coordinates $( 5 , k )$.
The line $A B$ has equation $3 x + 4 y = 7$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $k = - 2$.
\item Hence find the coordinates of the mid-point of $A B$.
\end{enumerate}\item Find the gradient of $A B$.
\item The line $A C$ is perpendicular to the line $A B$.
\begin{enumerate}[label=(\roman*)]
\item Find the gradient of $A C$.
\item Hence find an equation of the line $A C$.
\item Given that the point $C$ lies on the $x$-axis, find its $x$-coordinate.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2006 Q2 [10]}}