| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Equation of line through two points |
| Difficulty | Easy -1.2 This is a straightforward two-part question testing basic coordinate geometry skills: finding a line equation through two points using gradient formula, then applying the distance formula. Both are routine P1 techniques requiring minimal problem-solving, making this easier than average for A-level. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d Circles: equation (x-a)^2+(y-b)^2=r^21.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(C = (4, -2)\) | B1 | |
| \(m_{AB} = -1/2 \rightarrow m_{CD} = 2\) | M1 | Use of \(m_1 m_2 = -1\) on their \(m_{AB}\) |
| Equation of \(CD\) is \(y + 2 = 2(x - 4)\) oe | M1 | Use of *their* \(C\) and \(m_{CD}\) in a line equation |
| \(y = 2x - 10\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(AD^2 = (14-0)^2 + (-7-(-10))^2\) | M1 | Use *their* \(D\) in a correct method |
| \(AD = 14.3\) or \(\sqrt{205}\) | A1 [2] |
## Question 4:
### Part (i):
$C = (4, -2)$ | **B1** |
$m_{AB} = -1/2 \rightarrow m_{CD} = 2$ | **M1** | Use of $m_1 m_2 = -1$ on their $m_{AB}$
Equation of $CD$ is $y + 2 = 2(x - 4)$ oe | **M1** | Use of *their* $C$ and $m_{CD}$ in a line equation
$y = 2x - 10$ | **A1** [4] |
### Part (ii):
$AD^2 = (14-0)^2 + (-7-(-10))^2$ | **M1** | Use *their* $D$ in a correct method
$AD = 14.3$ or $\sqrt{205}$ | **A1** [2] |
---(i) Find the equation of the line $C D$, giving your answer in the form $y = m x + c$.\\
(ii) Find the distance $A D$.
\hfill \mbox{\textit{CAIE P1 2016 Q4 [6]}}