| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 11 |
| 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 multi-part question testing basic coordinate geometry skills: substituting a point into a line equation, finding gradients, using perpendicular gradient relationship (m₁m₂ = -1), and calculating distance. All parts are routine C1 techniques with no problem-solving or insight required, making it easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.10f Distance between points: using position vectors |
| Answer | Marks |
|---|---|
| \((8 - 3 - k = 0)\) so \(k = \underline{5}\) | B1 |
| (1 mark total) |
| Answer | Marks | Guidance |
|---|---|---|
| \(2y = 3x + k\); \(y = \dfrac{3}{2}x + \ldots\) and so \(m = \dfrac{3}{2}\) o.e. | M1, A1 | M1 for attempt to rearrange to \(y=\ldots\); A1 for clear statement that gradient is 1.5 |
| (2 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Perpendicular gradient \(= -\dfrac{2}{3}\) | B1ft | For using the perpendicular gradient rule correctly on their "1.5" |
| Equation of line: \(y - 4 = -\dfrac{2}{3}(x-1)\) | M1, A1ft | M1 for attempt at finding equation through \(A\) using their gradient; 1st A1ft for correct equation following through their changed gradient |
| \(3y + 2x - 14 = 0\) o.e. | A1 | As printed or equivalent with integer coefficients; allow \(3y+2x=14\) or \(3y=14-2x\) |
| (4 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 0 \Rightarrow B(7, 0)\) or \(x = 7\) | M1, A1ft | M1 for use of \(y=0\) to find \(x=\ldots\) in their equation; A1ft for \(x=7\) or \(-\dfrac{c}{a}\) |
| (2 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB^2 = (7-1)^2 + (4-0)^2\) | M1 | For attempt to find \(AB\) or \(AB^2\) |
| \(AB = \sqrt{52}\) or \(2\sqrt{13}\) | A1 | For any correct surd form; need not be simplified |
| (2 marks total) |
## Question 9:
### Part (a):
| $(8 - 3 - k = 0)$ so $k = \underline{5}$ | B1 | |
|(1 mark total)|||
### Part (b):
| $2y = 3x + k$; $y = \dfrac{3}{2}x + \ldots$ and so $m = \dfrac{3}{2}$ o.e. | M1, A1 | M1 for attempt to rearrange to $y=\ldots$; A1 for clear statement that gradient is 1.5 |
|(2 marks total)|||
### Part (c):
| Perpendicular gradient $= -\dfrac{2}{3}$ | B1ft | For using the perpendicular gradient rule correctly on their "1.5" |
| Equation of line: $y - 4 = -\dfrac{2}{3}(x-1)$ | M1, A1ft | M1 for attempt at finding equation through $A$ using their gradient; 1st A1ft for correct equation following through their changed gradient |
| $3y + 2x - 14 = 0$ o.e. | A1 | As printed or equivalent with integer coefficients; allow $3y+2x=14$ or $3y=14-2x$ |
|(4 marks total)|||
### Part (d):
| $y = 0 \Rightarrow B(7, 0)$ or $x = 7$ | M1, A1ft | M1 for use of $y=0$ to find $x=\ldots$ in their equation; A1ft for $x=7$ or $-\dfrac{c}{a}$ |
|(2 marks total)|||
### Part (e):
| $AB^2 = (7-1)^2 + (4-0)^2$ | M1 | For attempt to find $AB$ or $AB^2$ |
| $AB = \sqrt{52}$ or $2\sqrt{13}$ | A1 | For any correct surd form; need not be simplified |
|(2 marks total)|||9. The line $L _ { 1 }$ has equation $2 y - 3 x - k = 0$, where $k$ is a constant.
Given that the point $A ( 1,4 )$ lies on $L _ { 1 }$, find
\begin{enumerate}[label=(\alph*)]
\item the value of $k$,
\item the gradient of $L _ { 1 }$.
The line $L _ { 2 }$ passes through $A$ and is perpendicular to $L _ { 1 }$.
\item Find an equation of $L _ { 2 }$ giving your answer 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 $B$.
\item Find the coordinates of $B$.
\item Find the exact length of $A B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2011 Q9 [11]}}