| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| 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 coordinate geometry problem requiring finding perpendicular gradients (AB ⊥ BC), using parallel line properties, and solving simultaneous equations. While it involves multiple steps and connecting several concepts (perpendicularity condition, parallel lines, horizontal line), these are standard P1 techniques with straightforward algebraic manipulation. Slightly above average due to the multi-step nature and need to coordinate multiple geometric constraints, but no novel insight required. |
| 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 | Marks | Guidance |
| Gradient of \(AB = -\frac{1}{2} \rightarrow\) Gradient of \(BC = 2\) | M1 | Use of \(m_1 m_2 = -1\) for correct lines |
| Forms equation in \(h\): \(\frac{3h-2}{h} = 2\) | M1 | Uses normal line equation or gradients for \(h\) |
| \(h = 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vectors \(\overrightarrow{AB} \cdot \overrightarrow{BC} = 0\) | M1 | Use of vectors AB and BC |
| Solving | M1 | |
| \(h = 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of Pythagoras to find 3 lengths | M1 | |
| Solving | M1 | |
| \(h = 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y\) coordinate of \(D\) is \(6\), (\(3 \times\) 'their' \(h\)); \(\frac{6-0}{x-4} = 2 \rightarrow x = 7 \rightarrow D(7,6)\) | B1 | FT |
| Vectors: \(\overrightarrow{AD} \cdot \overrightarrow{AB} = 0\) | M1 A1 | Must use \(y = 6\). Realises the \(y\) values of \(C\) and \(D\) are equal. Uses gradient or line equation to find \(x\) |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient of $AB = -\frac{1}{2} \rightarrow$ Gradient of $BC = 2$ | M1 | Use of $m_1 m_2 = -1$ for correct lines |
| Forms equation in $h$: $\frac{3h-2}{h} = 2$ | M1 | Uses normal line equation or gradients for $h$ |
| $h = 2$ | A1 | |
**Alternative method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vectors $\overrightarrow{AB} \cdot \overrightarrow{BC} = 0$ | M1 | Use of vectors AB and BC |
| Solving | M1 | |
| $h = 2$ | A1 | |
**Alternative method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of Pythagoras to find 3 lengths | M1 | |
| Solving | M1 | |
| $h = 2$ | A1 | |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y$ coordinate of $D$ is $6$, ($3 \times$ 'their' $h$); $\frac{6-0}{x-4} = 2 \rightarrow x = 7 \rightarrow D(7,6)$ | B1 | FT |
| Vectors: $\overrightarrow{AD} \cdot \overrightarrow{AB} = 0$ | M1 A1 | Must use $y = 6$. Realises the $y$ values of $C$ and $D$ are equal. Uses gradient or line equation to find $x$ |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{ebf16cae-1e80-44d2-9c51-630f5dc3c11f-06_625_750_260_699}
The diagram shows a trapezium $A B C D$ in which the coordinates of $A , B$ and $C$ are (4, 0), (0, 2) and $( h , 3 h )$ respectively. The lines $B C$ and $A D$ are parallel, angle $A B C = 90 ^ { \circ }$ and $C D$ is parallel to the $x$-axis.\\
(i) Find, by calculation, the value of $h$.\\
(ii) Hence find the coordinates of $D$.\\
\hfill \mbox{\textit{CAIE P1 2019 Q4 [6]}}