| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Sector and arc length |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic coordinate geometry (finding intercepts, perpendicular lines) and standard circle formulas (arc length, sector area). Part (b) is scaffolded as 'show that', and part (c) is direct application of s=rθ. The only mild challenge is part (d) combining sector and triangle areas, but this is routine bookwork for P1 level. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y\) coordinate \(= 12\) | B1 | Check diagram; answer in main solution takes precedence over diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Gradient of \(l_1 = -\frac{3}{4}\) | B1 | May be implied by further work; sight of gradient \(\frac{4}{3}\) in \(l_2\) equation also scores; cannot be awarded from rearranged \(l_1\) alone |
| Gradient of \(l_2 = \frac{4}{3} \Rightarrow (y-6)=\frac{4}{3}(x-8)\) | M1 | Must attempt perpendicular gradient and find equation of \(l_2\); if using \(y=mx+c\) must proceed to \(c=\ldots\) |
| \(y\) coordinate \(= -\frac{14}{3}\) | A1* cso | Must be clearly stated as \(y\) coordinate; no errors after correct equation for \(l_2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Radius \(= \text{"12"} + \frac{14}{3} = \frac{50}{3}\) | B1ft | Follow through on answer to (a); may be implied by arc length |
| Length of arc \(= \frac{50}{3} \times 1.8 = 30\) | M1A1cao | \(\theta = 1.8\) with their \(r\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area of sector \(= \frac{1}{2}\times\left(\frac{50}{3}\right)^2\times 1.8\ (= 250)\) | M1 | Using \(\theta=1.8\) and their \(r\) |
| \(\text{"250"} + \frac{1}{2}\times\frac{50}{3}\times 8 = \text{"250"} + \frac{200}{3}\) | M1 | Adds sector area with correct method for triangle area; alternatives: \(CD=\sqrt{6^2+8^2}=10\), \(DE=\sqrt{8^2+\left(\frac{32}{3}\right)^2}=\frac{40}{3}\), Area\(=\frac{1}{2}\times10\times\frac{40}{3}=\frac{200}{3}\); or shoelace method |
| \(= \frac{950}{3}\) (units²) | A1cao | Accept \(316\frac{2}{3}\) or \(316.\dot{6}\) but not \(316.7\) |
# Question 7:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y$ coordinate $= 12$ | B1 | Check diagram; answer in main solution takes precedence over diagram |
**(1 mark)**
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Gradient of $l_1 = -\frac{3}{4}$ | B1 | May be implied by further work; sight of gradient $\frac{4}{3}$ in $l_2$ equation also scores; cannot be awarded from rearranged $l_1$ alone |
| Gradient of $l_2 = \frac{4}{3} \Rightarrow (y-6)=\frac{4}{3}(x-8)$ | M1 | Must attempt perpendicular gradient and find equation of $l_2$; if using $y=mx+c$ must proceed to $c=\ldots$ |
| $y$ coordinate $= -\frac{14}{3}$ | A1* cso | Must be clearly stated as $y$ coordinate; no errors after correct equation for $l_2$ |
**(3 marks)**
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Radius $= \text{"12"} + \frac{14}{3} = \frac{50}{3}$ | B1ft | Follow through on answer to (a); may be implied by arc length |
| Length of arc $= \frac{50}{3} \times 1.8 = 30$ | M1A1cao | $\theta = 1.8$ with their $r$ |
**(3 marks)**
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Area of sector $= \frac{1}{2}\times\left(\frac{50}{3}\right)^2\times 1.8\ (= 250)$ | M1 | Using $\theta=1.8$ and their $r$ |
| $\text{"250"} + \frac{1}{2}\times\frac{50}{3}\times 8 = \text{"250"} + \frac{200}{3}$ | M1 | Adds sector area with correct method for triangle area; alternatives: $CD=\sqrt{6^2+8^2}=10$, $DE=\sqrt{8^2+\left(\frac{32}{3}\right)^2}=\frac{40}{3}$, Area$=\frac{1}{2}\times10\times\frac{40}{3}=\frac{200}{3}$; or shoelace method |
| $= \frac{950}{3}$ (units²) | A1cao | Accept $316\frac{2}{3}$ or $316.\dot{6}$ but not $316.7$ |
**(3 marks, 10 marks total)**
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{877d03f2-d62c-4060-bdd2-f0d5dfbe6203-22_775_837_251_557}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The line $l _ { 1 }$ has equation $4 y + 3 x = 48$\\
The line $l _ { 1 }$ cuts the $y$-axis at the point $C$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item State the $y$ coordinate of $C$.
The point $D ( 8,6 )$ lies on $l _ { 1 }$\\
The line $l _ { 2 }$ passes through $D$ and is perpendicular to $l _ { 1 }$
The line $l _ { 2 }$ cuts the $y$-axis at the point $E$ as shown in Figure 3.
\item Show that the $y$ coordinate of $E$ is $- \frac { 14 } { 3 }$
A sector $B C E$ of a circle with centre $C$ is also shown in Figure 3.
Given that angle $B C E$ is 1.8 radians,
\item find the length of arc $B E$.
The region $C B E D$, shown shaded in Figure 3, consists of the sector $B C E$ joined to the triangle $C D E$.
\item Calculate the exact area of the region $C B E D$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2021 Q7 [10]}}