| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Midpoint of line segment |
| Difficulty | Easy -1.2 This is a straightforward two-part question testing basic coordinate geometry: finding a midpoint using the standard formula, then finding a perpendicular line equation. Both are routine procedures requiring only direct application of formulas with no problem-solving insight or multi-step reasoning beyond basic algebraic manipulation. |
| 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/Working | Marks | Guidance |
| \((3\tfrac{1}{2}, 2)\) | B1 | |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(m = \frac{-1-5}{5-2} = -2\) | B1 | |
| \(y - 6 = \frac{-1}{m}(x-8)\) | M1 | Use of \(m_1m_2 = -1\) and \(y-k = m(x-h)\) |
| \(x - 2y + 4 = 0\) | A1 | Accept any form |
| [3] |
## Question 2:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(3\tfrac{1}{2}, 2)$ | B1 | |
| **[1]** | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $m = \frac{-1-5}{5-2} = -2$ | B1 | |
| $y - 6 = \frac{-1}{m}(x-8)$ | M1 | Use of $m_1m_2 = -1$ and $y-k = m(x-h)$ |
| $x - 2y + 4 = 0$ | A1 | Accept any form |
| **[3]** | | |
---2 Points $A , B$ and $C$ have coordinates $( 2,5 ) , ( 5 , - 1 )$ and $( 8,6 )$ respectively.\\
(i) Find the coordinates of the mid-point of $A B$.\\
(ii) Find the equation of the line through $C$ perpendicular to $A B$. Give your answer in the form $a x + b y + c = 0$.
\hfill \mbox{\textit{CAIE P1 2010 Q2 [4]}}