| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | November |
| Marks | 5 |
| 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 coordinate geometry question requiring substitution of points into a line equation to find constants, followed by applying the standard midpoint formula. Both parts involve routine algebraic manipulation with no problem-solving insight needed, making it easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(32 - 4k = 20 \Rightarrow k = 3\) | M1A1 | Sub \((8, -4)\); [alt: \((2b+4)/(b-8) = -4/k\)] |
| \(4b + 3 \times 2b = 20\) | M1 | Sub \((b, 2b)\), \(4b + 2bk = 20\); M1 both M1 solving A1 |
| \(b = 2\) | A1 | A1 ] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mid-point \(= (5, 0)\) | B1✓ | Ft on *their* \(b\) |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $32 - 4k = 20 \Rightarrow k = 3$ | **M1A1** | Sub $(8, -4)$; [alt: $(2b+4)/(b-8) = -4/k$] |
| $4b + 3 \times 2b = 20$ | **M1** | Sub $(b, 2b)$, $4b + 2bk = 20$; **M1** both **M1** solving **A1** |
| $b = 2$ | **A1** | **A1** ] |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mid-point $= (5, 0)$ | **B1**✓ | Ft on *their* $b$ |
---4 The line $4 x + k y = 20$ passes through the points $A ( 8 , - 4 )$ and $B ( b , 2 b )$, where $k$ and $b$ are constants.\\
(i) Find the values of $k$ and $b$.\\
(ii) Find the coordinates of the mid-point of $A B$.
\hfill \mbox{\textit{CAIE P1 2014 Q4 [5]}}