| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 8 |
| 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 requiring standard techniques: rearranging to find gradient, using perpendicular gradient rule (m₁m₂ = -1), finding intercepts, and calculating triangle area. All steps are routine C1 procedures with no problem-solving insight needed, making it easier than average but not trivial due to multiple connected parts. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((m=)\ \dfrac{2}{3}\) | B1 | For \(\dfrac{2}{3}\) seen. Do not award for \(\dfrac{2}{3}x\); must be in part (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(B: (0, 4)\) | B1 | For coordinates of \(B\). Accept 4 marked on \(y\)-axis |
| Gradient: \(\dfrac{-1}{m} = -\dfrac{3}{2}\) | M1 | Use of perpendicular gradient rule. Follow through their \(m\) |
| \(y - 4 = -\dfrac{3x}{2}\) or equiv. e.g. \(y = -\dfrac{3x}{2}+4\), \(3x+2y-8=0\) | A1 | Correct equation any form, need not be simplified. Answer only 3/3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A: (-6, 0)\) | B1 | Coordinates of \(A\). Accept \(-6\) marked on \(x\)-axis |
| \(C:\ \dfrac{3x}{2} = 4 \Rightarrow x = \dfrac{8}{3}\) | B1ft | Coordinates of \(C\). Accept \(AC = \dfrac{26}{3}\). Accept \(x=\dfrac{8}{3}\) marked on \(x\)-axis. Follow through from \(l_2\) if \(>0\) |
| Area using \(\dfrac{1}{2}(x_c - x_A)y_B\) | M1 | Expression for area of triangle (all lengths \(> 0\)). Ft their 4, \(-6\) and \(\dfrac{8}{3}\) |
| \(= \dfrac{1}{2}\left(\dfrac{8}{3}+6\right)4 = \dfrac{52}{3}\left(= 17\tfrac{1}{3}\right)\) | A1cso | \(\dfrac{52}{3}\) or exact equivalent as single fraction or \(17\tfrac{1}{3}\) or \(17\tfrac{2}{6}\) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(BC = \dfrac{4}{6}\sqrt{52}\) | 2nd B1ft | From similar triangles or using \(C\) |
| Area using \(\dfrac{1}{2}(AB \times BC)\), \(AB = \sqrt{6^2+4^2} = \sqrt{52}\) | M1 | |
| \(= \dfrac{1}{2}\times\sqrt{52}\times\left(\dfrac{2}{3}\sqrt{52}\right) = \dfrac{52}{3}\left(=17\tfrac{1}{3}\right)\) | A1 |
## Question 6:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(m=)\ \dfrac{2}{3}$ | B1 | For $\dfrac{2}{3}$ seen. Do not award for $\dfrac{2}{3}x$; must be in part (a) |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $B: (0, 4)$ | B1 | For coordinates of $B$. Accept 4 marked on $y$-axis |
| Gradient: $\dfrac{-1}{m} = -\dfrac{3}{2}$ | M1 | Use of perpendicular gradient rule. Follow through their $m$ |
| $y - 4 = -\dfrac{3x}{2}$ or equiv. e.g. $y = -\dfrac{3x}{2}+4$, $3x+2y-8=0$ | A1 | Correct equation any form, need not be simplified. Answer only 3/3 |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A: (-6, 0)$ | B1 | Coordinates of $A$. Accept $-6$ marked on $x$-axis |
| $C:\ \dfrac{3x}{2} = 4 \Rightarrow x = \dfrac{8}{3}$ | B1ft | Coordinates of $C$. Accept $AC = \dfrac{26}{3}$. Accept $x=\dfrac{8}{3}$ marked on $x$-axis. Follow through from $l_2$ if $>0$ |
| Area using $\dfrac{1}{2}(x_c - x_A)y_B$ | M1 | Expression for area of triangle (all lengths $> 0$). Ft their 4, $-6$ and $\dfrac{8}{3}$ |
| $= \dfrac{1}{2}\left(\dfrac{8}{3}+6\right)4 = \dfrac{52}{3}\left(= 17\tfrac{1}{3}\right)$ | A1cso | $\dfrac{52}{3}$ or exact equivalent as single fraction or $17\tfrac{1}{3}$ or $17\tfrac{2}{6}$ etc |
**ALT:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $BC = \dfrac{4}{6}\sqrt{52}$ | 2nd B1ft | From similar triangles or using $C$ |
| Area using $\dfrac{1}{2}(AB \times BC)$, $AB = \sqrt{6^2+4^2} = \sqrt{52}$ | M1 | |
| $= \dfrac{1}{2}\times\sqrt{52}\times\left(\dfrac{2}{3}\sqrt{52}\right) = \dfrac{52}{3}\left(=17\tfrac{1}{3}\right)$ | A1 | |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ff1cdb91-0286-4bc8-9e67-451500b2bf74-07_647_927_274_513}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The line $l _ { 1 }$ has equation $2 x - 3 y + 12 = 0$
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of $l _ { 1 }$.
The line $l _ { 1 }$ crosses the $x$-axis at the point $A$ and the $y$-axis at the point $B$, as shown in Figure 1.
The line $l _ { 2 }$ is perpendicular to $l _ { 1 }$ and passes through $B$.
\item Find an equation of $l _ { 2 }$.
The line $l _ { 2 }$ crosses the $x$-axis at the point $C$.
\item Find the area of triangle $A B C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2012 Q6 [8]}}