| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Rectangle or parallelogram vertices |
| Difficulty | Moderate -0.3 This is a structured multi-part question testing standard coordinate geometry skills: finding gradients, parallel line equations, intercepts, and parallelogram area. While it requires multiple steps and combining several techniques, each individual part uses routine C1 methods with no novel problem-solving required. The parallelogram area calculation is slightly more involved than basic exercises, bringing it just slightly below average difficulty. |
| 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 |
| \(\frac{5}{4}\) | B1 | Exact equivalents such as 1.25 accepted, but not \(\frac{5}{4}x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \frac{5}{4}x + c\) | M1 | Uses line with parallel gradient \(\frac{5}{4}\) or gradient from part (a) |
| \(12, 5 \Rightarrow 5 = \frac{5}{4} \times 12 + c \Rightarrow c = \ldots\) | M1 | Method of finding equation of line with numerical gradient passing through \((12, 5)\) |
| \(y = \frac{5}{4}x - 10\) | A1 | Correct equation. Allow \(-\frac{40}{4}\) for \(-10\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(B = (0, -10)\) | B1ft | Follow through on their \(c\). Allow if \(-10\) marked correctly on diagram. Allow \(x=0, y=-10\) |
| \(C = (8, 0)\) | B1 | Correct coordinates. Allow if 8 marked correctly on diagram. Allow \(y=0, x=8\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Way 1: Area of Parallelogram \(= 3 + {'}10{'} \times {'}8{'}\) | M1 | Uses area of parallelogram \(= bh = 3+{'}10{'} \times {'}8{'}\); follow through on their 10 and their 8 |
| \(= 104\) | A1 | cao |
| Way 2: Trapezium \(AOCD\) + Triangle \(OCB\) \(= \frac{1}{2}(3+3+{'}10{'}) \times {'}8{'} + \frac{1}{2} \times {'}8{'} \times {'}10{'}\) | M1 | A correct method using their values for \(AOCD + OCB\) |
| \(= 104\) | A1 | cao |
| Way 3: 2 Triangles + Rectangle \(= 2 \times \frac{1}{2} \times {'}8{'} \times {'}10{'} + {'}8{'} \times 3\) | M1 | A correct method using their values for \(2 \times OBC\) + rectangle |
| \(= 104\) | A1 | cao |
| Way 4: Triangle \(ACD\) + Triangle \(ACB\) \(= 2 \times \frac{1}{2} \times {'}10{'} + 3 \times {'}8{'}\) | M1 | A correct method using their values for \(ACD + ABC\) |
| \(= 104\) | A1 | cao |
## Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{5}{4}$ | B1 | Exact equivalents such as 1.25 accepted, but **not** $\frac{5}{4}x$ |
## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{5}{4}x + c$ | M1 | Uses line with parallel gradient $\frac{5}{4}$ or gradient from part (a) |
| $12, 5 \Rightarrow 5 = \frac{5}{4} \times 12 + c \Rightarrow c = \ldots$ | M1 | Method of finding equation of line with numerical gradient passing through $(12, 5)$ |
| $y = \frac{5}{4}x - 10$ | A1 | Correct equation. Allow $-\frac{40}{4}$ for $-10$ |
## Question 8(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $B = (0, -10)$ | B1ft | Follow through on their $c$. Allow if $-10$ marked correctly on diagram. Allow $x=0, y=-10$ |
| $C = (8, 0)$ | B1 | Correct coordinates. Allow if 8 marked correctly on diagram. Allow $y=0, x=8$ |
## Question (d) — Area Calculation
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Way 1:** Area of Parallelogram $= 3 + {'}10{'} \times {'}8{'}$ | M1 | Uses area of parallelogram $= bh = 3+{'}10{'} \times {'}8{'}$; follow through on their 10 and their 8 |
| $= 104$ | A1 | cao |
| **Way 2:** Trapezium $AOCD$ + Triangle $OCB$ $= \frac{1}{2}(3+3+{'}10{'}) \times {'}8{'} + \frac{1}{2} \times {'}8{'} \times {'}10{'}$ | M1 | A correct method using their values for $AOCD + OCB$ |
| $= 104$ | A1 | cao |
| **Way 3:** 2 Triangles + Rectangle $= 2 \times \frac{1}{2} \times {'}8{'} \times {'}10{'} + {'}8{'} \times 3$ | M1 | A correct method using their values for $2 \times OBC$ + rectangle |
| $= 104$ | A1 | cao |
| **Way 4:** Triangle $ACD$ + Triangle $ACB$ $= 2 \times \frac{1}{2} \times {'}10{'} + 3 \times {'}8{'}$ | M1 | A correct method using their values for $ACD + ABC$ |
| $= 104$ | A1 | cao |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{937246f9-2b6a-48df-b919-c6db3d6f863b-20_1063_1319_251_365}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the straight line $l _ { 1 }$ with equation $4 y = 5 x + 12$
\begin{enumerate}[label=(\alph*)]
\item State the gradient of $l _ { 1 }$
The line $l _ { 2 }$ is parallel to $l _ { 1 }$ and passes through the point $E ( 12,5 )$, as shown in Figure 2.
\item Find the equation of $l _ { 2 }$. Write your answer in the form $y = m x + c$, where $m$ and $c$ are constants to be determined.
The line $l _ { 2 }$ cuts the $x$-axis at the point $C$ and the $y$-axis at the point $B$.
\item Find the coordinates of
\begin{enumerate}[label=(\roman*)]
\item the point $B$,
\item the point $C$.
The line $l _ { 1 }$ cuts the $y$-axis at the point $A$.\\
The point $D$ lies on $l _ { 1 }$ such that $A B C D$ is a parallelogram, as shown in Figure 2.
\end{enumerate}\item Find the area of $A B C D$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2018 Q8 [8]}}