| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Moderate -0.8 This is a straightforward C1 coordinate geometry question requiring standard techniques: finding perpendicular gradient (negative reciprocal), using point-slope form, rearranging to general form, finding intercepts, and calculating triangle area. All steps are routine textbook exercises with no problem-solving insight required, making it easier than average but not trivial due to multiple parts. |
| 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 of \(l_2\) is \(\frac{1}{2}\) or \(0.5\) or \(\frac{-1}{-2}\) | B1 | Must have \(\frac{1}{2}\) or \(0.5\) or \(\frac{-1}{-2}\); stated and stops, or used in line equation |
| \(y - 6 = \frac{1}{2}(x-5)\) or \(y = \frac{1}{2}x + c\) with \(c=\frac{7}{2}\) | M1 | Full method to obtain equation of line through \((5,6)\) with their \(m\) |
| \(x - 2y + 7 = 0\) or \(-x + 2y - 7 = 0\) or \(k(x-2y+7)=0\) with \(k\) an integer | A1 | Accept any multiple of correct equation with integer coefficients and \(= 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Puts \(x=0\) or \(y=0\) in their equation | M1 | Either one of the \(x\) or \(y\) coordinates using their equation |
| \(x\)-coordinate of \(A\) is \(-7\), \(y\)-coordinate of \(B\) is \(\frac{7}{2}\) | A1 cao | Needs both correct values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(OAB = \frac{1}{2}(7)\left(\frac{7}{2}\right) = \frac{49}{4}\) | M1 | Applies \(\pm\frac{1}{2}(\text{base})(\text{height})\) |
| \(= \frac{49}{4}\) (units)\(^2\) | A1 cso | Any exact equivalent e.g. \(12.25\); negative final answer is A0 but replacing by positive is A1 |
## Question 5:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of $l_2$ is $\frac{1}{2}$ or $0.5$ or $\frac{-1}{-2}$ | B1 | Must have $\frac{1}{2}$ or $0.5$ or $\frac{-1}{-2}$; stated and stops, or used in line equation |
| $y - 6 = \frac{1}{2}(x-5)$ or $y = \frac{1}{2}x + c$ with $c=\frac{7}{2}$ | M1 | Full method to obtain equation of line through $(5,6)$ with their $m$ |
| $x - 2y + 7 = 0$ or $-x + 2y - 7 = 0$ or $k(x-2y+7)=0$ with $k$ an integer | A1 | Accept any multiple of correct equation with integer coefficients and $= 0$ |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Puts $x=0$ or $y=0$ in their equation | M1 | Either one of the $x$ or $y$ coordinates using their equation |
| $x$-coordinate of $A$ is $-7$, $y$-coordinate of $B$ is $\frac{7}{2}$ | A1 cao | Needs both correct values |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $OAB = \frac{1}{2}(7)\left(\frac{7}{2}\right) = \frac{49}{4}$ | M1 | Applies $\pm\frac{1}{2}(\text{base})(\text{height})$ |
| $= \frac{49}{4}$ (units)$^2$ | A1 cso | Any exact equivalent e.g. $12.25$; negative final answer is A0 but replacing by positive is A1 |
---5. The line $l _ { 1 }$ has equation $y = - 2 x + 3$
The line $l _ { 2 }$ is perpendicular to $l _ { 1 }$ and passes through the point $( 5,6 )$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for $l _ { 2 }$ in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
The line $l _ { 2 }$ crosses the $x$-axis at the point $A$ and the $y$-axis at the point $B$.
\item Find the $x$-coordinate of $A$ and the $y$-coordinate of $B$.
Given that $O$ is the origin,
\item find the area of the triangle $O A B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2013 Q5 [7]}}