| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2002 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Rectangle or parallelogram vertices |
| Difficulty | Moderate -0.3 This is a multi-part coordinate geometry question requiring perpendicular gradients, simultaneous equations, and distance formula. While it involves several steps (finding perpendicular line, intersection point, calculating distances), each technique is standard and the rectangle structure provides clear guidance. Slightly easier than average due to straightforward application of routine methods without requiring geometric insight or problem-solving creativity. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| (i) m of AB \(= -2\). m of BC \(= -1\) (m) \(= \frac{1}{2}\). equation of BC \(y - 6 = \frac{1}{2}(x - 1)\) or \(2y = x + 11\) | B1, M1, DM1, A1∨ | Correct only; Used correctly; Correct formula needed to be used; A∨ mark for any correct equation |
| (ii) Sim eqns \(y = x - 1\) and answer above. Solution C (13,12) | M1, A1 | Correct method; Correct only |
| (iii) \(AB = \sqrt{20}\) and \(BC = \sqrt{180}\). perimeter \(= 2\sqrt{20} + 2\sqrt{180} = 35.8\) or \(35.7\) or \(16\sqrt{5}\) or \(\sqrt{1280}\) | M1, DM1, A1 | Use of Pythagoras once – \(\sqrt{20}\) ok; Use of \(2a + 2b\) – with Pythagoras twice; Correct only |
**(i)** m of AB $= -2$. m of BC $= -1$ (m) $= \frac{1}{2}$. equation of BC $y - 6 = \frac{1}{2}(x - 1)$ or $2y = x + 11$ | B1, M1, DM1, A1∨ | Correct only; Used correctly; Correct formula needed to be used; A∨ mark for any correct equation
**(ii)** Sim eqns $y = x - 1$ and answer above. Solution C (13,12) | M1, A1 | Correct method; Correct only
**(iii)** $AB = \sqrt{20}$ and $BC = \sqrt{180}$. perimeter $= 2\sqrt{20} + 2\sqrt{180} = 35.8$ or $35.7$ or $16\sqrt{5}$ or $\sqrt{1280}$ | M1, DM1, A1 | Use of Pythagoras once – $\sqrt{20}$ ok; Use of $2a + 2b$ – with Pythagoras twice; Correct only9\\
\includegraphics[max width=\textwidth, alt={}, center]{a10ad459-6f86-4845-ba28-e4e394bf3d1e-4_719_958_264_589}
The diagram shows a rectangle $A B C D$, where $A$ is $( 3,2 )$ and $B$ is $( 1,6 )$.\\
(i) Find the equation of $B C$.
Given that the equation of $A C$ is $y = x - 1$, find\\
(ii) the coordinates of $C$,\\
(iii) the perimeter of the rectangle $A B C D$.
\hfill \mbox{\textit{CAIE P1 2002 Q9 [9]}}