| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Rectangle or parallelogram vertices |
| Difficulty | Standard +0.3 This is a straightforward coordinate geometry question requiring finding a parallel line equation, calculating intercepts, using the parallelogram property that opposite sides are equal vectors, and finding area using the standard formula. All techniques are routine P1/AS-level material with no novel insight required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempts gradient of \(3x-4y+20=0 \Rightarrow y = \frac{3}{4}x+5\) | M1 | Attempt to rearrange and make \(y\) subject; expect \(\pm 4y = \ldots\) followed by \(y = \ldots\) or equivalent |
| Equation \(l_2\): \(y - 0 = \frac{3}{4}(x-8) \Rightarrow 3x-4y-24=0\) | M1, A1 | M1 for using same gradient as \(l_1\) with \(R(8,0)\) to form linear equation; A1 for \(3x-4y-24=0\) or integer multiple. Correct equation implies full marks (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P = \left(-\frac{20}{3}, 0\right)\), \(Q = (0,5)\) | B1 | May be implied by subsequent calculations or seen on diagram |
| Area \(PQRS = PR \times OQ = \left(8 + \frac{20}{3}\right) \times 5 = \frac{220}{3}\) | M1, A1 | Full attempt at area of parallelogram using their \(P\) and \(Q\); candidates may use Shoelace algorithm or determinant method; A1 for \(\frac{220}{3}\) (condone awrt 73.3). SC: sign slip on \(Q\) giving \(P=(-\frac{20}{3},0)\), \(Q=(0,-5)\) scores 011 (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\overrightarrow{QR} = \begin{pmatrix} 8 \\ -5 \end{pmatrix} \Rightarrow S = \left(-\frac{20}{3}+8, 0-5\right) = \left(\frac{4}{3}, -5\right)\) | M1, A1 | M1 for full attempt to find both coordinates of \(S\) (via \(\overrightarrow{OP}+\overrightarrow{QR}\), or solving equations, or equating lengths, or rotational symmetry); A1 for \(\left(\frac{4}{3},-5\right)\), allow \(x=\ldots, y=\ldots\), condone awrt \((1.33, -5)\). SC: \(S=\left(\frac{44}{3},5\right)\) scores SC10 (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Solve \(y=\frac{3}{4}x-6\) with \(y=-\frac{5}{8}\left(x+\frac{20}{3}\right)\): \(\frac{3}{4}x-6=-\frac{5}{8}\left(x+\frac{20}{3}\right) \Rightarrow \frac{11}{8}x = \frac{11}{6} \Rightarrow x = \ldots\) | M1 | |
| \(S = \left(\frac{4}{3}, -5\right)\) | A1 | (2 marks) |
## Question 6:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempts gradient of $3x-4y+20=0 \Rightarrow y = \frac{3}{4}x+5$ | M1 | Attempt to rearrange and make $y$ subject; expect $\pm 4y = \ldots$ followed by $y = \ldots$ or equivalent |
| Equation $l_2$: $y - 0 = \frac{3}{4}(x-8) \Rightarrow 3x-4y-24=0$ | M1, A1 | M1 for using same gradient as $l_1$ with $R(8,0)$ to form linear equation; A1 for $3x-4y-24=0$ or integer multiple. **Correct equation implies full marks** **(3 marks)** |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P = \left(-\frac{20}{3}, 0\right)$, $Q = (0,5)$ | B1 | May be implied by subsequent calculations or seen on diagram |
| Area $PQRS = PR \times OQ = \left(8 + \frac{20}{3}\right) \times 5 = \frac{220}{3}$ | M1, A1 | Full attempt at area of parallelogram using their $P$ and $Q$; candidates may use Shoelace algorithm or determinant method; A1 for $\frac{220}{3}$ (condone awrt 73.3). SC: sign slip on $Q$ giving $P=(-\frac{20}{3},0)$, $Q=(0,-5)$ scores 011 **(3 marks)** |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\overrightarrow{QR} = \begin{pmatrix} 8 \\ -5 \end{pmatrix} \Rightarrow S = \left(-\frac{20}{3}+8, 0-5\right) = \left(\frac{4}{3}, -5\right)$ | M1, A1 | M1 for full attempt to find both coordinates of $S$ (via $\overrightarrow{OP}+\overrightarrow{QR}$, or solving equations, or equating lengths, or rotational symmetry); A1 for $\left(\frac{4}{3},-5\right)$, allow $x=\ldots, y=\ldots$, condone awrt $(1.33, -5)$. SC: $S=\left(\frac{44}{3},5\right)$ scores SC10 **(2 marks)** |
**Alt(c):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Solve $y=\frac{3}{4}x-6$ with $y=-\frac{5}{8}\left(x+\frac{20}{3}\right)$: $\frac{3}{4}x-6=-\frac{5}{8}\left(x+\frac{20}{3}\right) \Rightarrow \frac{11}{8}x = \frac{11}{6} \Rightarrow x = \ldots$ | M1 | |
| $S = \left(\frac{4}{3}, -5\right)$ | A1 | **(2 marks)** |6. The line $l _ { 1 }$ has equation $3 x - 4 y + 20 = 0$
The line $l _ { 2 }$ cuts the $x$-axis at $R ( 8,0 )$ and is parallel to $l _ { 1 }$
\begin{enumerate}[label=(\alph*)]
\item Find the equation of $l _ { 2 }$, writing your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers to be found.
The line $l _ { 1 }$ cuts the $x$-axis at $P$ and the $y$-axis at $Q$.\\
Given that $P Q R S$ is a parallelogram, find
\item the area of $P Q R S$,
\item the coordinates of $S$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2020 Q6 [8]}}