| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Rectangle or parallelogram vertices |
| Difficulty | Moderate -0.3 This is a straightforward coordinate geometry question requiring finding equations of lines (using two points and perpendicular gradients) and solving simultaneous equations. While it involves multiple steps, each technique is standard GCSE/AS-level material with no novel insight required. The kite context provides clear structure, making it slightly easier than average A-level questions. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Eqn of \(AC\): \(y = -\frac{1}{2}x + 4\) (gradient must be \(\Delta y / \Delta x\)) | M1A1 | Uses gradient and a given point for equation. CAO |
| Gradient of \(OB = 2 \rightarrow y = 2x\) (If \(y\) missing only penalise once) | M1 A1 | Use of \(m_1 m_2 = -1\), answers only ok |
| Answer | Marks | Guidance |
|---|---|---|
| Simultaneous equations \(\rightarrow ((1.6, 3.2))\) | M1 | Equate and solve for M1 and reach \(\geq 1\) solution |
| This is mid-point of \(OB \rightarrow B\ (3.2, 6.4)\) | M1 A1 | Uses mid-point. CAO |
| or: Let coordinates of \(B\ (h,k)\); \(OA = AB \rightarrow h^2 = 8k - k^2\); \(OC = BC \rightarrow k^2 = 16h - h^2 \rightarrow (3.2, 6.4)\) | M1 for both equations, M1 for solving with \(y = 2x\) | |
| or: gradients \((\frac{k-4}{h} \times \frac{k}{h-8} = -1)\) | M1 for gradient product as \(-1\), M1 solving with \(y = 2x\) | |
| or: Pythagoras: \(h^2 + (k-4)^2 + (h-8)^2 + k^2 = 4^2 + 8^2\) | M1 for complete equation, M1 solving with \(y = 2x\) |
## Question 5(i):
| Eqn of $AC$: $y = -\frac{1}{2}x + 4$ (gradient must be $\Delta y / \Delta x$) | M1A1 | Uses gradient and a given point for equation. CAO |
|---|---|---|
| Gradient of $OB = 2 \rightarrow y = 2x$ (If $y$ missing only penalise once) | M1 A1 | Use of $m_1 m_2 = -1$, answers only ok |
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## Question 5(ii):
| Simultaneous equations $\rightarrow ((1.6, 3.2))$ | M1 | Equate and solve for M1 and reach $\geq 1$ solution |
|---|---|---|
| This is mid-point of $OB \rightarrow B\ (3.2, 6.4)$ | M1 A1 | Uses mid-point. CAO |
| **or:** Let coordinates of $B\ (h,k)$; $OA = AB \rightarrow h^2 = 8k - k^2$; $OC = BC \rightarrow k^2 = 16h - h^2 \rightarrow (3.2, 6.4)$ | | M1 for both equations, M1 for solving with $y = 2x$ |
| **or:** gradients $(\frac{k-4}{h} \times \frac{k}{h-8} = -1)$ | | M1 for gradient product as $-1$, M1 solving with $y = 2x$ |
| **or:** Pythagoras: $h^2 + (k-4)^2 + (h-8)^2 + k^2 = 4^2 + 8^2$ | | M1 for complete equation, M1 solving with $y = 2x$ |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{5df7bd9f-31cc-41a3-b1c0-3ee9366e6d8a-08_558_785_258_680}
The diagram shows a kite $O A B C$ in which $A C$ is the line of symmetry. The coordinates of $A$ and $C$ are $( 0,4 )$ and $( 8,0 )$ respectively and $O$ is the origin.\\
(i) Find the equations of $A C$ and $O B$.\\
(ii) Find, by calculation, the coordinates of $B$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q5 [7]}}