| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Parallel line through point |
| Difficulty | Moderate -0.8 This is a straightforward coordinate geometry question requiring standard techniques: finding a line equation from two points, using parallel line properties, applying distance formula, and calculating triangle area. All steps are routine P1 material with no novel insight required, making it easier than average but not trivial due to the multi-part nature and algebraic manipulation needed. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03c Straight line models: in variety of contexts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(m = \frac{7-2}{15-4} \left(= \frac{5}{11}\right)\) | M1 | Correct method for the gradient |
| \(y - 2 = \frac{5}{11}(x-4)\) or \(y - 7 = \frac{5}{11}(x-15)\) or \(y = \frac{5}{11}x + c \Rightarrow 2 = \frac{5}{11}\times4 + c \Rightarrow c = ...\left(\frac{2}{11}\right)\) | M1 | Correct straight line method using their gradient and point \(A\) or \(B\) with \(x\) and \(y\) correctly placed. Note \(\frac{y-2}{7-2} = \frac{x-4}{15-4}\) scores both M marks |
| \(5x - 11y + 2 = 0\) | A1 | Correct equation in required form including "\(=0\)". Allow any integer multiple |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = mx + c \Rightarrow \begin{cases} 2 = 4m+c \\ 7 = 15m+c \end{cases}\) | M1 | Substitutes \(x=4, y=2\) and \(x=15, y=7\) into \(y=mx+c\) to obtain 2 equations |
| \(m = ...\left(\frac{5}{11}\right),\ c = ...\left(\frac{2}{11}\right)\) | M1 | Solves 2 equations of the form \(\alpha = m\beta + c\) |
| \(5x - 11y + 2 = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((15-x)^2 + (7-2)^2 = (5\sqrt{5})^2\) or \(\sqrt{(15-x)^2 + (7-2)^2} = 5\sqrt{5}\) or \((5\sqrt{5})^2 = 5^2 + CD^2\) | M1 | Correct application of Pythagoras for triangle \(BCD\) |
| Way 1: \((15-x)^2 + 25 = 125 \Rightarrow (15-x)^2 = 100 \Rightarrow x = ...\) or Way 2: \(125 = 25 + CD^2 \Rightarrow CD^2 = 100 \Rightarrow CD = 10 \Rightarrow x = ...\) | M1 | Complete, correct and full method to obtain at least one \(x\) coordinate for \(C\) |
| \((5, 2)\) or \((25, 2)\) | A1 | At least one correct position for \(C\) |
| \((5, 2)\) and \((25, 2)\) | A1 | Both correct pairs of coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area \(= \frac{1}{2}("5" - 4) \times 5\) | M1 | Applies correct triangle area formula for one of their points of the form \((k, 2),\ k \neq 15\). This mark is for e.g. \(\frac{1}{2}(k-4)\times5\) where \(k\) is one of their calculated \(x\) coordinates from part (b), or \(\frac{1}{2}(11 \pm CD)\times 5\) where \(CD\) is their calculated value |
| \(= \frac{5}{2}\) | A1 | Correct minimum area. Must be exactly \(2.5\), not awrt \(2.5\) |
## Question 9:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $m = \frac{7-2}{15-4} \left(= \frac{5}{11}\right)$ | M1 | Correct method for the gradient |
| $y - 2 = \frac{5}{11}(x-4)$ or $y - 7 = \frac{5}{11}(x-15)$ or $y = \frac{5}{11}x + c \Rightarrow 2 = \frac{5}{11}\times4 + c \Rightarrow c = ...\left(\frac{2}{11}\right)$ | M1 | Correct straight line method using their gradient and point $A$ or $B$ with $x$ and $y$ correctly placed. Note $\frac{y-2}{7-2} = \frac{x-4}{15-4}$ scores both M marks |
| $5x - 11y + 2 = 0$ | A1 | Correct equation in required form including "$=0$". Allow any integer multiple |
**ALT:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = mx + c \Rightarrow \begin{cases} 2 = 4m+c \\ 7 = 15m+c \end{cases}$ | M1 | Substitutes $x=4, y=2$ and $x=15, y=7$ into $y=mx+c$ to obtain 2 equations |
| $m = ...\left(\frac{5}{11}\right),\ c = ...\left(\frac{2}{11}\right)$ | M1 | Solves 2 equations of the form $\alpha = m\beta + c$ |
| $5x - 11y + 2 = 0$ | A1 | |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(15-x)^2 + (7-2)^2 = (5\sqrt{5})^2$ or $\sqrt{(15-x)^2 + (7-2)^2} = 5\sqrt{5}$ or $(5\sqrt{5})^2 = 5^2 + CD^2$ | M1 | Correct application of Pythagoras for triangle $BCD$ |
| Way 1: $(15-x)^2 + 25 = 125 \Rightarrow (15-x)^2 = 100 \Rightarrow x = ...$ or Way 2: $125 = 25 + CD^2 \Rightarrow CD^2 = 100 \Rightarrow CD = 10 \Rightarrow x = ...$ | M1 | Complete, correct and full method to obtain at least one $x$ coordinate for $C$ |
| $(5, 2)$ **or** $(25, 2)$ | A1 | At least one correct position for $C$ |
| $(5, 2)$ **and** $(25, 2)$ | A1 | Both correct pairs of coordinates |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Area $= \frac{1}{2}("5" - 4) \times 5$ | M1 | Applies correct triangle area formula for one of their points of the form $(k, 2),\ k \neq 15$. This mark is for e.g. $\frac{1}{2}(k-4)\times5$ where $k$ is one of their calculated $x$ coordinates from part (b), or $\frac{1}{2}(11 \pm CD)\times 5$ where $CD$ is their calculated value |
| $= \frac{5}{2}$ | A1 | Correct minimum area. Must be exactly $2.5$, not awrt $2.5$ |\begin{enumerate}
\item Given that
\end{enumerate}
\begin{itemize}
\item the point $A$ has coordinates $( 4,2 )$
\item the point $B$ has coordinates $( 15,7 )$
\item the line $l _ { 1 }$ passes through $A$ and $B$\\
(a) find an equation for $l _ { 1 }$, giving your answer in the form $p x + q y + r = 0$ where $p , q$ and $r$ are integers to be found.
\end{itemize}
The line $l _ { 2 }$ passes through $A$ and is parallel to the $x$-axis.\\
The point $C$ lies on $l _ { 2 }$ so that the length of $B C$ is $5 \sqrt { 5 }$\\
(b) Find both possible pairs of coordinates of the point $C$.\\
(c) Hence find the minimum possible area of triangle $A B C$.
\hfill \mbox{\textit{Edexcel P1 2024 Q9 [9]}}