| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Standard +0.3 This is a straightforward coordinate geometry question requiring finding a perpendicular line (using negative reciprocal gradients), finding intersection points by substituting x=0 or y=0, then calculating distances and area. All steps are routine applications of standard techniques 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient \(AB = -\frac{2}{5}\) | B1 | States gradient of \(AB\) is \(-\frac{2}{5}\) |
| \(y\) coordinate of \(A\) is \(2\) | B1 | States \(y\) coordinate of \(A = 2\) |
| Uses perpendicular gradients \(y = +\frac{5}{2}x + c\) | M1 | Uses form \(y = mx + c\) with \(m\) = their adapted \(-\frac{2}{5}\) and \(c\) = their 2; or \(y - y_1 = m(x-x_1)\) with \((x_1,y_1)=(0,2)\) |
| \(\Rightarrow 2y - 5x = 4\) | A1* | Proceeds to given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses Pythagoras to find \(AB\) or \(AD\): either \(\sqrt{5^2+2^2}\) or \(\sqrt{\left(\frac{4}{5}\right)^2+2^2}\) | M1 | Look for \(\sqrt{5^2+2^2}\) or \(\sqrt{\left(\frac{4}{5}\right)^2+2^2}\); alternatively finds lengths \(BD\) and \(AO\), look for \(\left(5+\frac{4}{5}\right)\) and \(2\) |
| Uses area \(ABCD = AD \times AB = \sqrt{29} \times \sqrt{\frac{116}{25}}\) | M1 | Full method for area of rectangle \(ABCD\); alternatively \(2\times\frac{1}{2}BD\times AO = 2\times\frac{1}{2}\text{'5.8'}\times\text{'2'}\) |
| Area \(ABCD = 11.6\) | A1 | \(11.6\) or exact equivalent \(\frac{58}{5}\) |
# Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient $AB = -\frac{2}{5}$ | B1 | States gradient of $AB$ is $-\frac{2}{5}$ |
| $y$ coordinate of $A$ is $2$ | B1 | States $y$ coordinate of $A = 2$ |
| Uses perpendicular gradients $y = +\frac{5}{2}x + c$ | M1 | Uses form $y = mx + c$ with $m$ = their adapted $-\frac{2}{5}$ and $c$ = their 2; or $y - y_1 = m(x-x_1)$ with $(x_1,y_1)=(0,2)$ |
| $\Rightarrow 2y - 5x = 4$ | A1* | Proceeds to given answer |
# Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses Pythagoras to find $AB$ or $AD$: either $\sqrt{5^2+2^2}$ or $\sqrt{\left(\frac{4}{5}\right)^2+2^2}$ | M1 | Look for $\sqrt{5^2+2^2}$ or $\sqrt{\left(\frac{4}{5}\right)^2+2^2}$; alternatively finds lengths $BD$ and $AO$, look for $\left(5+\frac{4}{5}\right)$ and $2$ |
| Uses area $ABCD = AD \times AB = \sqrt{29} \times \sqrt{\frac{116}{25}}$ | M1 | Full method for area of rectangle $ABCD$; alternatively $2\times\frac{1}{2}BD\times AO = 2\times\frac{1}{2}\text{'5.8'}\times\text{'2'}$ |
| Area $ABCD = 11.6$ | A1 | $11.6$ or exact equivalent $\frac{58}{5}$ |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a75c9ef7-b648-47be-bad1-fc8b315be3df-10_602_999_260_534}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a rectangle $A B C D$.\\
The point $A$ lies on the $y$-axis and the points $B$ and $D$ lie on the $x$-axis as shown in Figure 1. Given that the straight line through the points $A$ and $B$ has equation $5 y + 2 x = 10$
\begin{enumerate}[label=(\alph*)]
\item show that the straight line through the points $A$ and $D$ has equation $2 y - 5 x = 4$
\item find the area of the rectangle $A B C D$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 Q8 [7]}}