| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2009 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Rectangle or parallelogram vertices |
| Difficulty | Standard +0.8 This is a multi-part coordinate geometry problem requiring understanding of rectangle properties (perpendicular sides, diagonals bisect), midpoint theorem, and perpendicular gradient relationships. Part (i) is straightforward, but parts (ii)-(iv) require setting up and solving equations using the condition that AD ⊥ CD (product of gradients = -1), which demands solid problem-solving beyond routine exercises. The multi-step nature and need to connect several geometric properties elevates this above average difficulty. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.05g Exact trigonometric values: for standard angles1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (i) y-coordinate same as the y-coordinate of the mid-point of \(AC\). | B1 [1] | co |
| Answer | Marks | Guidance |
|---|---|---|
| nb \(AC = 20\), \(M(6, 6)\) \(MD = 10 \rightarrow D(16, 6)\) and \(B(-4, 6)\) | M1 A1, A1 [3] | any use of y-step ÷ x-step for M mark. co |
| Answer | Marks | Guidance |
|---|---|---|
| or Pyth \(h^2 + 8^2 + 8^2 + (12-h)^2 = 400\) | M1, M1, DM1A1 [4] | Used correctly with the two gradients. Forming a quadratic equation. Solution of equation. co |
| (iv) Area \(= \sqrt{320} \times \sqrt{80} \rightarrow 160\) | M1 M1, A1 [3] | M1 for method for one of the lengths. M1 for base × height. co |
**(i)** y-coordinate same as the y-coordinate of the mid-point of $AC$. | B1 [1] | co
**(ii)** $m$ of $AD = \frac{8}{h}$ or $\frac{h-12}{-8}$
$m$ of $CD = \frac{12-h}{8}$ or $\frac{-h}{8}$
nb $AC = 20$, $M(6, 6)$ $MD = 10 \rightarrow D(16, 6)$ and $B(-4, 6)$ | M1 A1, A1 [3] | any use of y-step ÷ x-step for M mark. co
**(iii)** Product of gradients $= -1 \rightarrow h^2 - 12h - 64 = 0 \rightarrow h = 16$ or $-4$
so $x_D = 16$ and $x_B = -4$
or Pyth $h^2 + 8^2 + 8^2 + (12-h)^2 = 400$ | M1, M1, DM1A1 [4] | Used correctly with the two gradients. Forming a quadratic equation. Solution of equation. co
**(iv)** Area $= \sqrt{320} \times \sqrt{80} \rightarrow 160$ | M1 M1, A1 [3] | M1 for method for one of the lengths. M1 for base × height. co
(or Area $= 2 \times$ area of a triangle with base $= BD, \rightarrow 2 \times \frac{1}{2} \times 20 \times 8 = 160$)
(or matrix method)
**Total: [11]**
---9\\
\includegraphics[max width=\textwidth, alt={}, center]{b566719c-216e-41e5-8431-da77e1dad73e-3_554_723_1557_712}
The diagram shows a rectangle $A B C D$. The point $A$ is $( 0 , - 2 )$ and $C$ is $( 12,14 )$. The diagonal $B D$ is parallel to the $x$-axis.\\
(i) Explain why the $y$-coordinate of $D$ is 6 .
The $x$-coordinate of $D$ is $h$.\\
(ii) Express the gradients of $A D$ and $C D$ in terms of $h$.\\
(iii) Calculate the $x$-coordinates of $D$ and $B$.\\
(iv) Calculate the area of the rectangle $A B C D$.
\hfill \mbox{\textit{CAIE P1 2009 Q9 [11]}}