| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Rectangle or parallelogram vertices |
| Difficulty | Standard +0.3 This is a straightforward coordinate geometry problem requiring perpendicular gradient property (m₁ × m₂ = -1) and solving simultaneous equations. While it involves multiple steps, each technique is standard A-level fare with no novel insight needed—slightly easier than average due to the routine application of well-practiced methods. |
| 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 | Marks | Guidance |
| Gradient of \(AB = 2\) | B1 | co |
| Gradient of \(BC = -\frac{1}{2}\) | M1 | For use of \(m_1 m_2 = -1\) |
| \(\rightarrow\) Eqn of \(BC\) is \(y - 11 = -\frac{1}{2}(x-5)\) | A1 | co, unsimplified is fine |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient of \(AC\) (or \(AX\)) is \(\frac{1}{3}\) | B1 | co |
| \(\rightarrow\) eqn of \(AC\) is \(y - 3 = \frac{1}{3}(x-1)\) | M1 | Correct form of line equation + sim eqns |
| or \(y - 4 = \frac{1}{3}(x-4)\) | co | |
| Sim equations \(\rightarrow C(13, 7)\) | A1 | answer only \(-0/3\), assumed \(AB = BC\); uses graph or table and gets exactly \((13,7)\): allow the 3 marks for (ii) |
| [3] |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of $AB = 2$ | B1 | co |
| Gradient of $BC = -\frac{1}{2}$ | M1 | For use of $m_1 m_2 = -1$ |
| $\rightarrow$ Eqn of $BC$ is $y - 11 = -\frac{1}{2}(x-5)$ | A1 | co, unsimplified is fine |
| **[3]** | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of $AC$ (or $AX$) is $\frac{1}{3}$ | B1 | co |
| $\rightarrow$ eqn of $AC$ is $y - 3 = \frac{1}{3}(x-1)$ | M1 | Correct form of line equation + sim eqns |
| or $y - 4 = \frac{1}{3}(x-4)$ | | co |
| Sim equations $\rightarrow C(13, 7)$ | A1 | answer only $-0/3$, assumed $AB = BC$; uses graph or table and gets exactly $(13,7)$: allow the 3 marks for (ii) |
| **[3]** | | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{11bfe5bd-604c-43e5-81e7-4c1f5676bcbb-3_602_751_255_696}
The diagram shows a triangle $A B C$ in which $A$ has coordinates ( 1,3 ), $B$ has coordinates ( 5,11 ) and angle $A B C$ is $90 ^ { \circ }$. The point $X ( 4,4 )$ lies on $A C$. Find\\
(i) the equation of $B C$,\\
(ii) the coordinates of $C$.
\hfill \mbox{\textit{CAIE P1 2012 Q5 [6]}}