| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular from point to line |
| Difficulty | Standard +0.3 This is a straightforward coordinate geometry question requiring standard techniques: finding the perpendicular line equation, solving simultaneous equations for intersection point X, using reflection in a line for point D, and calculating distances. While multi-part with several steps, each component uses routine P1 methods without requiring novel insight or complex problem-solving. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Gradient of \(AC = -\frac{1}{2}\) | B1 | Correct gradient. |
| Perpendicular gradient \(= 2\) | M1 | Use of \(m_1m_2 = -1\) |
| Eqn of \(BX\) is \(y - 2 = 2(x - 2)\) | M1 | Correct form of equation |
| Sim Eqns \(2y + x = 16\) with \(y = 2x - 2\) | A1 | co [4] |
| \(\rightarrow (4, 6)\) | ||
| (ii) \(X\) is mid-point of \(BD\), \(D\) is \((6, 10)\) | M1 A1√ | Any valid method. It on (i). [2] |
| (iii) \(AB = \sqrt{(4^2 + 2^2)} = \sqrt{200}\) | M1 | Use of Pythagoras once. |
| \(BC = \sqrt{(2^2 + 6^2)} = \sqrt{40}\) | M1 | 4 lengths added. |
| \(\rightarrow\) Perimeter \(= 2\sqrt{200} + 2\sqrt{40}\) | DM1 | 4 lengths added. |
| \(\rightarrow\) Perimeter \(= 40.9\) | A1 | co [3] |
**(i)** Gradient of $AC = -\frac{1}{2}$ | B1 | Correct gradient.
Perpendicular gradient $= 2$ | M1 | Use of $m_1m_2 = -1$
Eqn of $BX$ is $y - 2 = 2(x - 2)$ | M1 | Correct form of equation
Sim Eqns $2y + x = 16$ with $y = 2x - 2$ | A1 | co [4]
$\rightarrow (4, 6)$ |
**(ii)** $X$ is mid-point of $BD$, $D$ is $(6, 10)$ | M1 A1√ | Any valid method. It on (i). [2]
**(iii)** $AB = \sqrt{(4^2 + 2^2)} = \sqrt{200}$ | M1 | Use of Pythagoras once.
$BC = \sqrt{(2^2 + 6^2)} = \sqrt{40}$ | M1 | 4 lengths added.
$\rightarrow$ Perimeter $= 2\sqrt{200} + 2\sqrt{40}$ | DM1 | 4 lengths added.
$\rightarrow$ Perimeter $= 40.9$ | A1 | co [3]
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**DM1 for quadratic:** Quadratic must be set to 0. Factors. Attempt at two brackets. Each bracket set to 0 and solved. Formula. Correct formula. Correct use, but allow for numerical slips in $b^2$ and $-4ac$.11\\
\includegraphics[max width=\textwidth, alt={}, center]{d71002bb-b6f0-42a3-89fb-f2769d5c3779-4_563_965_813_591}
In the diagram, the points $A$ and $C$ lie on the $x$ - and $y$-axes respectively and the equation of $A C$ is $2 y + x = 16$. The point $B$ has coordinates ( 2,2 ). The perpendicular from $B$ to $A C$ meets $A C$ at the point $X$.\\
(i) Find the coordinates of $X$.
The point $D$ is such that the quadrilateral $A B C D$ has $A C$ as a line of symmetry.\\
(ii) Find the coordinates of $D$.\\
(iii) Find, correct to 1 decimal place, the perimeter of $A B C D$.
\hfill \mbox{\textit{CAIE P1 2008 Q11 [9]}}