| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Coordinates from geometric constraints |
| Difficulty | Standard +0.3 This is a multi-part coordinate geometry question requiring gradient calculations, perpendicular line conditions, and rectangle properties. While it has multiple steps (3 parts, likely 7-8 marks total), each part uses standard C1 techniques: finding line equations from two points, using perpendicular gradient property (m₁m₂ = -1), and applying rectangle diagonal properties. No novel insight required—just systematic application of core methods. Slightly easier than average due to straightforward setup and clear geometric constraints. |
| 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 |
| \(\text{gradient} = \frac{y_1-y_2}{x_1-x_2} = \frac{2-(-4)}{-1-7} = -\frac{3}{4}\) | M1, A1 | M1: uses gradient formula with points \(L\) and \(M\), attempt to substitute correct numbers. Formula implied by correct \(\frac{2-(-4)}{-1-7}\). A1: any correct single fraction gradient i.e. \(\frac{6}{-8}\) or equivalent |
| \(y - 2 = -\frac{3}{4}(x+1)\) or \(y+4 = -\frac{3}{4}(x-7)\) or \(y = \text{their}\,-\frac{3}{4}x + c\) | M1 | Uses their gradient with either \((-1,2)\) or \((7,-4)\) to form linear equation |
| \(\Rightarrow \pm(4y + 3x - 5) = 0\) | A1 | Accept \(\pm k(4y+3x-5)=0\) with \(k\) an integer (implies previous M1) |
| Method 3: substitute \(x=-1, y=2\) and \(x=7, y=-4\) into \(ax+by+c=0\): \(-a+2b+c=0\) and \(7a-4b+c=0\); solve to get \(a=3, b=4, c=-5\) or multiple | M1, A1, M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts gradient \(LM \times\) gradient \(MN = -1\): \(-\frac{3}{4}\times\frac{p+4}{16-7}=-1\) or \(\frac{p+4}{16-7}=\frac{4}{3}\) | M1 | Attempts to use gradient \(LM \times\) gradient \(MN = -1\); allow sign errors |
| \(p + 4 = \frac{9\times4}{3} \Rightarrow p = 8\) | M1, A1 | M1: attempt to solve linear equation in \(p\). A1: cao \(p=8\) |
| Alternative: Pythagoras: \((p+4)^2+9^2+(6^2+8^2)=(p-2)^2+17^2\) | M1 | Attempt Pythagoras correct way round (allow sign errors) |
| \(p^2+8p+16+81+36+64=p^2-4p+4+289 \Rightarrow p=...\) | M1 | |
| \(p = 8\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either \((y=) p+6\) or \(2+p+4\) | M1 | For using their numerical value of \(p\) and adding 6; may be done by any complete method (vectors, drawing, perpendicular lines through \(L\) and \(N\)). Assuming \(x=7\) is M0 |
| \((y=)\ 14\) | A1 | Accept 14 for both marks as long as no incorrect working seen (ignore LHS, allow \(k\)). If wrong working results fortuitously in 14, give M0A0. Allow \((8,14)\) as answer |
## Question 7:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{gradient} = \frac{y_1-y_2}{x_1-x_2} = \frac{2-(-4)}{-1-7} = -\frac{3}{4}$ | M1, A1 | M1: uses gradient formula with points $L$ and $M$, attempt to substitute correct numbers. Formula implied by correct $\frac{2-(-4)}{-1-7}$. A1: any correct single fraction gradient i.e. $\frac{6}{-8}$ or equivalent |
| $y - 2 = -\frac{3}{4}(x+1)$ or $y+4 = -\frac{3}{4}(x-7)$ or $y = \text{their}\,-\frac{3}{4}x + c$ | M1 | Uses their gradient with either $(-1,2)$ or $(7,-4)$ to form linear equation |
| $\Rightarrow \pm(4y + 3x - 5) = 0$ | A1 | Accept $\pm k(4y+3x-5)=0$ with $k$ an integer (implies previous M1) |
| Method 3: substitute $x=-1, y=2$ and $x=7, y=-4$ into $ax+by+c=0$: $-a+2b+c=0$ and $7a-4b+c=0$; solve to get $a=3, b=4, c=-5$ or multiple | M1, A1, M1, A1 | |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts gradient $LM \times$ gradient $MN = -1$: $-\frac{3}{4}\times\frac{p+4}{16-7}=-1$ or $\frac{p+4}{16-7}=\frac{4}{3}$ | M1 | Attempts to use gradient $LM \times$ gradient $MN = -1$; allow sign errors |
| $p + 4 = \frac{9\times4}{3} \Rightarrow p = 8$ | M1, A1 | M1: attempt to solve linear equation in $p$. A1: cao $p=8$ |
| **Alternative:** Pythagoras: $(p+4)^2+9^2+(6^2+8^2)=(p-2)^2+17^2$ | M1 | Attempt Pythagoras correct way round (allow sign errors) |
| $p^2+8p+16+81+36+64=p^2-4p+4+289 \Rightarrow p=...$ | M1 | |
| $p = 8$ | A1 | |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Either $(y=) p+6$ or $2+p+4$ | M1 | For using their numerical value of $p$ and adding 6; may be done by any complete method (vectors, drawing, perpendicular lines through $L$ and $N$). Assuming $x=7$ is M0 |
| $(y=)\ 14$ | A1 | Accept 14 for both marks as long as no incorrect working seen (ignore LHS, allow $k$). If wrong working results fortuitously in 14, give M0A0. Allow $(8,14)$ as answer |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6db8acbd-7f61-46ff-8fdc-f0f4a8363aa6-10_869_1073_267_440}
\captionsetup{labelformat=empty}
\caption{Diagram NOT to scale}
\end{center}
\end{figure}
Figure 2
Figure 2 shows a right angled triangle $L M N$.
The points $L$ and $M$ have coordinates ( $- 1,2$ ) and ( $7 , - 4$ ) respectively.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for the straight line passing through the points $L$ and $M$.
Give your answer in the form $a x + b y + c = 0$, where $a$, $b$ and $c$ are integers.
Given that the coordinates of point $N$ are ( $16 , p$ ), where $p$ is a constant, and angle $L M N = 90 ^ { \circ }$,
\item find the value of $p$.
Given that there is a point $K$ such that the points $L , M , N$, and $K$ form a rectangle,
\item find the $y$ coordinate of $K$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2014 Q7 [9]}}