| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | June |
| 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 C1 coordinate geometry question requiring standard techniques: finding gradient of AB, using perpendicular gradient property (negative reciprocal), writing equation of perpendicular line, finding y-intercept, and calculating triangle area using ½×base×height. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to multiple parts. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(AB: m = \frac{2-7}{8-6} = -\frac{5}{2}\) | B1 | Expression for gradient of \(AB\); does not need \(=-2.5\) |
| Using \(m_1m_2=-1\): \(m_2=\frac{2}{5}\) | M1 | Use of perpendicular gradient rule |
| \(y-7=\frac{2}{5}(x-6)\), giving \(2x-5y+23=0\) (integer coefficients) | M1, A1 | Use \((6,7)\) and changed gradient; correct equation with "\(=0\)" and integer coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Using \(x=0\): \(y=\frac{23}{5}\) or \(4.6\) | M1, A1ft | Substitute \(x=0\) into answer to (a); requires simplified fraction or exact decimal |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of triangle \(= \frac{1}{2}\times8\times\frac{23}{5} = \frac{92}{5}\) | M1 A1 | For \(\frac{1}{2}\times8\times y_C\); \(y\)-coordinate of \(C\) must be positive; accept \(18\frac{2}{5}\), \(18.4\), \(\frac{184}{10}\) |
# Question 8:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $AB: m = \frac{2-7}{8-6} = -\frac{5}{2}$ | B1 | Expression for gradient of $AB$; does not need $=-2.5$ |
| Using $m_1m_2=-1$: $m_2=\frac{2}{5}$ | M1 | Use of perpendicular gradient rule |
| $y-7=\frac{2}{5}(x-6)$, giving $2x-5y+23=0$ (integer coefficients) | M1, A1 | Use $(6,7)$ and changed gradient; correct equation with "$=0$" and integer coefficients |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Using $x=0$: $y=\frac{23}{5}$ or $4.6$ | M1, A1ft | Substitute $x=0$ into answer to (a); requires simplified fraction or exact decimal |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of triangle $= \frac{1}{2}\times8\times\frac{23}{5} = \frac{92}{5}$ | M1 A1 | For $\frac{1}{2}\times8\times y_C$; $y$-coordinate of $C$ must be positive; accept $18\frac{2}{5}$, $18.4$, $\frac{184}{10}$ |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e72d0d82-af0e-4f36-8446-a67b764fd7f3-09_908_1043_201_495}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The points $A$ and $B$ have coordinates $( 6,7 )$ and $( 8,2 )$ respectively.\\
The line $l$ passes through the point $A$ and is perpendicular to the line $A B$, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for $l$ in the form $a x + b y + c = 0$, where $a$, $b$ and $c$ are integers.
Given that $l$ intersects the $y$-axis at the point $C$, find
\item the coordinates of $C$,
\item the area of $\triangle O C B$, where $O$ is the origin.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2009 Q8 [8]}}