| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Moderate -0.3 This is a multi-part coordinate geometry question that is slightly easier than average. Part (a) requires simple reading of gradient and y-intercept from a given equation. Part (b) uses perpendicular gradients (standard technique) to find l₂. Part (c) requires calculating distances AB and BC using the distance formula, then multiplying for area. All techniques are routine A-level methods with no novel insight required, though the multi-step nature and exact arithmetic in part (c) prevents it from being significantly below average. |
| Spec | 1.02i Represent inequalities: graphically on coordinate plane1.03a 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 | 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 \(\frac{7}{2}\) |
| (2 marks) | ||
| (b) Perpendicular gradient \(l_2 = -\frac{2}{3}\) | B1ft | 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 \(l\). Eg. \(-\frac{1}{1.5} = \frac{y-...}{x-...}\) |
| Equation of line is: \(y - 5 = -\frac{2}{3}(x-1)\) | M1A1 | 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 \(\frac{y-5}{-\frac{2}{3}} = x - ... and 5 = -\frac{2}{3} \times 1 + c \Rightarrow c = \frac{17}{3}\) cso \(3y + 2x - 17 = 0\) |
| (4 marks) | ||
| (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 \(\frac{17.5}{x} = \frac{3}{2} \Rightarrow x = ...\). \(C= (8.5, 0)\). Allow equivalents such as \(x = 8.5\) at \(C\) |
| \(AB = \sqrt{(1-0)^2 + (5-3.5)^2} = \left(\frac{\sqrt{13}}{2}\right)\) | 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 | dM1A1 | 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 \(\frac{65}{4}\). |
| (5 marks) | ||
| (11 marks) |
(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 $\frac{7}{2}$
| | **(2 marks)**
(b) Perpendicular gradient $l_2 = -\frac{2}{3}$ | B1ft | 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 $l$. Eg. $-\frac{1}{1.5} = \frac{y-...}{x-...}$
Equation of line is: $y - 5 = -\frac{2}{3}(x-1)$ | M1A1 | 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 $\frac{y-5}{-\frac{2}{3}} = x - ... and 5 = -\frac{2}{3} \times 1 + c \Rightarrow c = \frac{17}{3}$ cso $3y + 2x - 17 = 0$
| | **(4 marks)**
(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 $\frac{17.5}{x} = \frac{3}{2} \Rightarrow x = ...$. $C= (8.5, 0)$. Allow equivalents such as $x = 8.5$ at $C$
$AB = \sqrt{(1-0)^2 + (5-3.5)^2} = \left(\frac{\sqrt{13}}{2}\right)$ | 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 | dM1A1 | 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 $\frac{65}{4}$.
| | **(5 marks)**
| | **(11 marks)**
---\includegraphics{figure_2}
The straight line $l_1$ has equation $2y = 3x + 7$
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 $A$. [2]
\end{enumerate}
\end{enumerate}
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 $\angle ABC = 90°$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find an equation of $l_2$ in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are integers. [4]
\end{enumerate}
The rectangle $ABCD$, shown shaded in Figure 2, has vertices at the points $A$, $B$, $C$ and $D$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the exact area of rectangle $ABCD$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q6 [11]}}