| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Coordinates from geometric constraints |
| Difficulty | Standard +0.3 This is a standard multi-part coordinate geometry question requiring gradient calculation, perpendicular lines (right angle condition), distance formula, triangle area, and line equation. All techniques are routine C1 content with straightforward application, though the multiple parts and algebraic manipulation (simplifying √80 to 4√5) elevate it slightly above average difficulty. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of \(AB = \frac{4}{8} = \frac{1}{2}\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of \(BC = -2\), \(\frac{4-2}{k-7} = -2\) | M1 | (or full Pythag. method) |
| \(k = 6\) | A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB = \sqrt{4^2 + 8^2}\) | ||
| \(= \sqrt{80} = \sqrt{16}\sqrt{5} = 4\sqrt{5}\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(BC = \sqrt{1^2 + 2^2} = \sqrt{5}\) (or \(AC = \sqrt{7^2 + 6^2} = \sqrt{85}\)) | B1 ft | |
| Area of \(ABC = \frac{1}{2}(4\sqrt{5} \times \sqrt{5}) = 10\) | M1 A1 | Other exact methods score M1 A2; non-exact methods score M1 A0 (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y - 2 = -2(x - 7)\) | B1 | |
| \(2x + y - 16 = 0\) | B1 | (2 marks) |
## Question 8:
### Part (a)
Gradient of $AB = \frac{4}{8} = \frac{1}{2}$ | M1 A1 | (2 marks)
### Part (b)
Gradient of $BC = -2$, $\frac{4-2}{k-7} = -2$ | M1 | (or full Pythag. method)
$k = 6$ | A1 | (2 marks)
### Part (c)
$AB = \sqrt{4^2 + 8^2}$ | |
$= \sqrt{80} = \sqrt{16}\sqrt{5} = 4\sqrt{5}$ | A1 | (3 marks)
### Part (d)
$BC = \sqrt{1^2 + 2^2} = \sqrt{5}$ (or $AC = \sqrt{7^2 + 6^2} = \sqrt{85}$) | B1 ft |
Area of $ABC = \frac{1}{2}(4\sqrt{5} \times \sqrt{5}) = 10$ | M1 A1 | Other exact methods score M1 A2; non-exact methods score M1 A0 (3 marks)
### Part (e)
$y - 2 = -2(x - 7)$ | B1 |
$2x + y - 16 = 0$ | B1 | (2 marks)
---8. The points $A ( - 1 , - 2 ) , B ( 7,2 )$ and $C ( k , 4 )$, where $k$ is a constant, are the vertices of $\triangle A B C$. Angle $A B C$ is a right angle.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of $A B$.
\item Calculate the value of $k$.
\item Show that the length of $A B$ may be written in the form $p \sqrt { 5 }$, where $p$ is an integer to be found.
\item Find the exact value of the area of $\triangle A B C$.
\item Find an equation for the straight line $l$ passing through $B$ and $C$. Give your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q8 [12]}}