| 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 C1 coordinate geometry question with multiple routine parts: finding gradients, using perpendicular gradient condition (m₁m₂ = -1), distance formula, triangle area, and line equations. While it has 5 parts totaling perhaps 10-12 marks, each step uses basic formulas with no novel insight required. Slightly above average difficulty due to length and the algebraic manipulation involved, but all techniques are standard textbook exercises. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks |
|---|---|
| Gradient of \(AB = \frac{2-(-2)}{7-(-1)} = \frac{4}{8} = \frac{1}{2}\) | M1 A1 |
| Answer | Marks |
|---|---|
| Gradient of \(BC = -2\) (perpendicular to \(AB\)) | M1 |
| \(\frac{4-2}{k-7} = -2 \Rightarrow 2 = -2(k-7) \Rightarrow k = 6\) | A1 |
| Answer | Marks |
|---|---|
| \(AB = \sqrt{(7-(-1))^2 + (2-(-2))^2} = \sqrt{64+16} = \sqrt{80} = 4\sqrt{5}\) | M1 A1 |
| So \(p = 4\) | A1 |
| Answer | Marks |
|---|---|
| \(BC = \sqrt{(6-7)^2+(4-2)^2} = \sqrt{1+4} = \sqrt{5}\) | M1 |
| Area \(= \frac{1}{2} \times 4\sqrt{5} \times \sqrt{5} = \frac{1}{2} \times 20 = 10\) | M1 A1 |
| Answer | Marks |
|---|---|
| Gradient of \(BC = -2\), through \(B(7,2)\): \(y-2 = -2(x-7)\) | M1 |
| \(2x + y - 16 = 0\) | A1 |
# Question 9:
## Part (a):
| Gradient of $AB = \frac{2-(-2)}{7-(-1)} = \frac{4}{8} = \frac{1}{2}$ | M1 A1 | |
## Part (b):
| Gradient of $BC = -2$ (perpendicular to $AB$) | M1 | |
| $\frac{4-2}{k-7} = -2 \Rightarrow 2 = -2(k-7) \Rightarrow k = 6$ | A1 | |
## Part (c):
| $AB = \sqrt{(7-(-1))^2 + (2-(-2))^2} = \sqrt{64+16} = \sqrt{80} = 4\sqrt{5}$ | M1 A1 | |
| So $p = 4$ | A1 | |
## Part (d):
| $BC = \sqrt{(6-7)^2+(4-2)^2} = \sqrt{1+4} = \sqrt{5}$ | M1 | |
| Area $= \frac{1}{2} \times 4\sqrt{5} \times \sqrt{5} = \frac{1}{2} \times 20 = 10$ | M1 A1 | |
## Part (e):
| Gradient of $BC = -2$, through $B(7,2)$: $y-2 = -2(x-7)$ | M1 | |
| $2x + y - 16 = 0$ | A1 | |9. 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 Q9 [12]}}