| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| 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 coordinate geometry question requiring standard techniques: finding a perpendicular gradient (-5/3), using point-slope form, then finding intersection points and calculating triangle area. Part (a) is a 'show that' which guides students to the answer. The multi-step nature adds slight complexity, but all steps are routine AS-level procedures with no problem-solving insight required. |
| 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 |
| Gradient of \(l_2\) is \(-\frac{5}{3}\) | B1 | 1.1b |
| Complete attempt to find equation of \(l_2\) using \(B(8,0)\) and changed gradient, e.g. \(y - 0 = -\frac{1}{m_1}(x-8)\) | M1 | 1.1b |
| \(5x + 3y = 40\) | A1* | 2.1 — Must show at least one intermediate line; clear work with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces \(A(-10, 0)\) | B1 | 2.2a |
| Attempts to solve \(y = \frac{3}{5}x + 6\) and \(5x + 3y = 40\) simultaneously to find \(y\)-coordinate of \(C\) | M1 | 1.1b |
| \(y\)-coordinate of \(C\) is \(\frac{135}{17}\) | A1 | 1.1b |
| Complete attempt at area \(ABC = \frac{1}{2} \times (8 + 10) \times \frac{135}{17}\) | dM1 | 2.1 |
| \(= \frac{1215}{17}\) | A1 | 1.1b |
## Question 10(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of $l_2$ is $-\frac{5}{3}$ | B1 | 1.1b |
| Complete attempt to find equation of $l_2$ using $B(8,0)$ and changed gradient, e.g. $y - 0 = -\frac{1}{m_1}(x-8)$ | M1 | 1.1b |
| $5x + 3y = 40$ | A1* | 2.1 — Must show at least one intermediate line; clear work with no errors seen |
---
## Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $A(-10, 0)$ | B1 | 2.2a |
| Attempts to solve $y = \frac{3}{5}x + 6$ and $5x + 3y = 40$ simultaneously to find $y$-coordinate of $C$ | M1 | 1.1b |
| $y$-coordinate of $C$ is $\frac{135}{17}$ | A1 | 1.1b |
| Complete attempt at area $ABC = \frac{1}{2} \times (8 + 10) \times \frac{135}{17}$ | dM1 | 2.1 |
| $= \frac{1215}{17}$ | A1 | 1.1b |10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ce4f8375-0d88-4e48-85de-35f7e90b014d-20_643_767_276_648}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
The line $l _ { 1 }$ has equation $y = \frac { 3 } { 5 } x + 6$\\
The line $l _ { 2 }$ is perpendicular to $l _ { 1 }$ and passes through the point $B ( 8,0 )$, as shown in the sketch in Figure 4.
\begin{enumerate}[label=(\alph*)]
\item Show that an equation for line $l _ { 2 }$ is
$$5 x + 3 y = 40$$
Given that
\begin{itemize}
\item lines $l _ { 1 }$ and $l _ { 2 }$ intersect at the point $C$
\item line $l _ { 1 }$ crosses the $x$-axis at the point $A$
\item find the exact area of triangle $A B C$, giving your answer as a fully simplified fraction in the form $\frac { p } { q }$
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2023 Q10 [8]}}