| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2006 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular from point to line |
| Difficulty | Moderate -0.3 This is a standard coordinate geometry question requiring finding the equation of a perpendicular line and then solving simultaneous equations to find an intersection point. The steps are routine: calculate gradient of AB, use perpendicular gradient property (-1/m), write equation of CD using point-slope form, then solve two linear equations. While it requires multiple steps, each is a textbook technique with no novel insight needed, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(m\) of \(AB = \frac{8}{12}\) \(m\) of perpendicular \(= -\frac{12}{8}\) eqn of \(CD\) \(y - 15 = -\frac{1}{2}(x - 6)\) | M1 M1 A1 [3] | Use of \(m_1m_2 = -1\) and \(y\)-step/\(x\)-step. Correct form of eqn of line. co. |
| (ii) eqn of \(AB\) \(y - 3 = \frac{2}{3}(x - 1)\) Sim eqns \(2y + 3x = 48\) and \(3y = 2x + 7\) \(\rightarrow D(10, 9)\) | M1 A1∨ DM1 A1 [4] | Could be given in (i). Needs both M marks from (i). co. |
(i) $m$ of $AB = \frac{8}{12}$ $m$ of perpendicular $= -\frac{12}{8}$ eqn of $CD$ $y - 15 = -\frac{1}{2}(x - 6)$ | M1 M1 A1 [3] | Use of $m_1m_2 = -1$ and $y$-step/$x$-step. Correct form of eqn of line. co.
(ii) eqn of $AB$ $y - 3 = \frac{2}{3}(x - 1)$ Sim eqns $2y + 3x = 48$ and $3y = 2x + 7$ $\rightarrow D(10, 9)$ | M1 A1∨ DM1 A1 [4] | Could be given in (i). Needs both M marks from (i). co.
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\includegraphics[max width=\textwidth, alt={}, center]{dd2cb0ec-5df9-4d99-9e15-5ae1f1c07b96-3_684_771_260_685}
The three points $A ( 1,3 ) , B ( 13,11 )$ and $C ( 6,15 )$ are shown in the diagram. The perpendicular from $C$ to $A B$ meets $A B$ at the point $D$. Find\\
(i) the equation of $C D$,\\
(ii) the coordinates of $D$.
\hfill \mbox{\textit{CAIE P1 2006 Q5 [7]}}