| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | January |
| Marks | 11 |
| 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 standard techniques: reading gradient from equation, finding intercepts, using perpendicular gradients (negative reciprocal), and calculating distance/area. Part (a) is trivial recall, part (b) is routine perpendicular line work, and part (c) requires Pythagoras for distances. Slightly above average only due to the multi-step nature and rectangle area calculation, but all techniques are standard C1 material with no novel insight required. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | (i) \(\frac{3}{2}\) or equivalents such as 1.5 | B1 |
| (ii) \((0, 3.5)\) Accept \(y = 3\frac{1}{2}\) | B1 | cao intercept \(= (0,3.5)\). Accept 3.5, \(y=3.5\) and equivalences such as \(\frac{7}{2}\). |
| (b) | Perpendicular gradient \(l_2 = -\frac{2}{3}\) | B1 ft |
| Equation of line is: \(y - 5 = -\frac{2}{3}(x-1)\) | M1 A1 | For an attempt at finding the equation of \(l_2\) using \((1,5)\) and their adapted gradient. Condone for this mark a gradient of \(\frac{3}{2}\) going to \(\frac{3}{5}\). Eg. Allow for \(\frac{y-5}{x-1} = \frac{2}{3}\). If the form \(y = mx + c\) is used it must be a full method to find \(c\) with \((1,5)\) and an adapted gradient. |
| \(3y + 2x - 17 = 0\) | A1 | For an correct unsimplified equation of the line through \((1,5)\) with the correct gradient. Allow \(\frac{y-5}{-2} = \frac{x-1}{3} \Rightarrow x = ...\)or \(5 = -\frac{2}{3} \times 1 + c \Rightarrow c = \frac{17}{3}\). cso \(3y + 2x - 17 = 0\). |
| (c) | Point C: \(y = 0 \Rightarrow 2x = 17 \Rightarrow x = 8.5\) oe | M1, A1 |
| \(AB = \sqrt{(1-0)^2 + (5-3.5)^2} = \left(\frac{\sqrt{13}}{2}\right)\) | M1 (either) | An attempt to use \(\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\) for \(AB\) or \(BC\). There is no need to "calculate" these. Evidence of an attempt would be \(AB^2 = 1^2 + 1.5^2 \Rightarrow AB = ...\). |
| \(BC = \sqrt{(8.5-1)^2 + (5-0)^2} = \left(\frac{\sqrt{325}}{2}\right)\) | ||
| Area rectangle \(= AB \times BC = \frac{\sqrt{13}}{2} \times \frac{\sqrt{325}}{2} = \frac{\sqrt{13} \times \sqrt{13 \times 25}}{4} = \frac{5 \times 13}{4} = 16.25\) oe | dM1 A1 | Multiplying together their values of \(AB\) and \(BC\) to find area \(ABCD\). It is dependent upon both M"s having been scored. cao 16.25 or equivalents such as \(\frac{65}{4}\). |
(a) | (i) $\frac{3}{2}$ or equivalents such as 1.5 | B1 | cao gradient $= 1.5$. Accept equivalences such as $\frac{3}{2}$.
| (ii) $(0, 3.5)$ Accept $y = 3\frac{1}{2}$ | B1 | cao intercept $= (0,3.5)$. Accept 3.5, $y=3.5$ and equivalences such as $\frac{7}{2}$.
(b) | Perpendicular gradient $l_2 = -\frac{2}{3}$ | B1 ft | For using the perpendicular gradient rule, $m = -\frac{1}{m_2}$ on their "1.5". Accept $-\frac{1}{1.5}$, or this as part of their equation for $l_2$. Eg. $-\frac{1}{1.5} = \frac{y-...}{x-...}$.
| Equation of line is: $y - 5 = -\frac{2}{3}(x-1)$ | M1 A1 | For an attempt at finding the equation of $l_2$ using $(1,5)$ and their adapted gradient. Condone for this mark a gradient of $\frac{3}{2}$ going to $\frac{3}{5}$. Eg. Allow for $\frac{y-5}{x-1} = \frac{2}{3}$. If the form $y = mx + c$ is used it must be a full method to find $c$ with $(1,5)$ and an adapted gradient.
| $3y + 2x - 17 = 0$ | A1 | For an correct unsimplified equation of the line through $(1,5)$ with the correct gradient. Allow $\frac{y-5}{-2} = \frac{x-1}{3} \Rightarrow x = ...$or $5 = -\frac{2}{3} \times 1 + c \Rightarrow c = \frac{17}{3}$. cso $3y + 2x - 17 = 0$.
(c) | Point C: $y = 0 \Rightarrow 2x = 17 \Rightarrow x = 8.5$ oe | M1, A1 | An attempt to use their equation found in part b to find the $x$ coordinate of $C$. They must either use the equation of $l_2$ and set $y = 0 \Rightarrow x = ...$ or use its gradient. $17.5 = \frac{3}{2} \Rightarrow x = ...$.
| $AB = \sqrt{(1-0)^2 + (5-3.5)^2} = \left(\frac{\sqrt{13}}{2}\right)$ | M1 (either) | An attempt to use $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$ for $AB$ or $BC$. There is no need to "calculate" these. Evidence of an attempt would be $AB^2 = 1^2 + 1.5^2 \Rightarrow AB = ...$.
| $BC = \sqrt{(8.5-1)^2 + (5-0)^2} = \left(\frac{\sqrt{325}}{2}\right)$ | | |
| Area rectangle $= AB \times BC = \frac{\sqrt{13}}{2} \times \frac{\sqrt{325}}{2} = \frac{\sqrt{13} \times \sqrt{13 \times 25}}{4} = \frac{5 \times 13}{4} = 16.25$ oe | dM1 A1 | Multiplying together their values of $AB$ and $BC$ to find area $ABCD$. It is dependent upon both M"s having been scored. cao 16.25 or equivalents such as $\frac{65}{4}$.
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6081d81b-51d2-4140-9834-71ef7fd700b0-12_650_885_255_603}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The straight line $l _ { 1 }$ has equation $2 y = 3 x + 7$\\
The line $l _ { 1 }$ crosses the $y$-axis at the point $A$ as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item State the gradient of $l _ { 1 }$
\item Write down the coordinates of the point $A$.
Another straight line $l _ { 2 }$ intersects $l _ { 1 }$ at the point $B ( 1,5 )$ and crosses the $x$-axis at the point $C$, as shown in Figure 2.
Given that $\angle A B C = 90 ^ { \circ }$,
\end{enumerate}\item find an equation of $l _ { 2 }$ in the form $a x + b y + c = 0$, where $a$, b and $c$ are integers.
The rectangle $A B C D$, shown shaded in Figure 2, has vertices at the points $A , B , C$ and $D$.
\item Find the exact area of rectangle $A B C D$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2014 Q6 [11]}}