| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Coordinates from geometric constraints |
| Difficulty | Standard +0.8 This question requires finding coordinates from multiple geometric constraints (perpendicular diagonals, midpoint condition), involving simultaneous equations with perpendicular gradients and the section formula. While systematic, it demands careful coordinate geometry manipulation across multiple steps and is more challenging than standard textbook exercises. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.10a Vectors in 2D: i,j notation and column vectors |
**(i)**
- B1: Gradient of $AC = \frac{1}{2}$. co
- M1: Gradient of $BD = -2$. Use of $m_1 m_2 = -1$ with $AC$
- M1: Eqn of $BD$ is $y - 6 = -2(x - 3)$. Correct formula for straight line
- M1: Eqn of $AC$ is $y + 1 = \frac{1}{2}(x + 1)$
- A1: Sim eqns $\Rightarrow M (5, 2)$. Solution
- M1: Vector move – or midpoint back. Correct method. $\checkmark$ on $M$
- A1: $\Rightarrow D (7, -2)$. co
**(ii)**
- M1: Ratio of $AM : MC = \sqrt{45} : \sqrt{20}$. Correct distance formula
- A1: or Vector step $\Rightarrow 3 : 2$. Looks at the two $x$ or $y$ steps. Must be numerical, $1.5$ ok, not as roots
---9\\
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The diagram shows a quadrilateral $A B C D$ in which the point $A$ is ( $- 1 , - 1$ ), the point $B$ is ( 3,6 ) and the point $C$ is (9,4). The diagonals $A C$ and $B D$ intersect at $M$. Angle $B M A = 90 ^ { \circ }$ and $B M = M D$. Calculate\\
(i) the coordinates of $M$ and $D$,\\
(ii) the ratio $A M : M C$.
\hfill \mbox{\textit{CAIE P1 2011 Q9 [9]}}