| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Parameter from distance condition |
| Difficulty | Moderate -0.3 This is a straightforward coordinate geometry question requiring standard application of the distance formula (solving a quadratic equation) and perpendicular gradient condition (m₁ × m₂ = -1). Both parts are routine textbook exercises with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 1.03b Straight lines: parallel and perpendicular relationships1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((9-p)^2 + (3p)^2 = 169\) | M1 | Or \(\sqrt{\phantom{x}} = 13\) |
| \(10p^2 - 18p - 88 (=0)\) oe | A1 | 3-term quad |
| \(p = 4\) or \(-11/5\) oe | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient of given line \(= -\frac{2}{3}\) | B1 | |
| Hence gradient of \(AB = \frac{3}{2}\) | M1 | Attempt using \(m_1 m_2 = -1\) |
| \(\frac{3}{2} = \frac{3p}{9-p}\) oe eg \(\left(\frac{-2}{3}\right)\!\left(\frac{3p}{9-p}\right) = 1\) (includes previous M1) | M1 | Or vectors \(\begin{pmatrix}9-p\\3p\end{pmatrix}\cdot\begin{pmatrix}3\\-2\end{pmatrix}\) |
| \(p = 3\) | A1 [4] |
## Question 7(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(9-p)^2 + (3p)^2 = 169$ | M1 | Or $\sqrt{\phantom{x}} = 13$ |
| $10p^2 - 18p - 88 (=0)$ oe | A1 | 3-term quad |
| $p = 4$ **or** $-11/5$ oe | A1 [3] | |
## Question 7(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of given line $= -\frac{2}{3}$ | B1 | |
| Hence gradient of $AB = \frac{3}{2}$ | M1 | Attempt using $m_1 m_2 = -1$ |
| $\frac{3}{2} = \frac{3p}{9-p}$ oe eg $\left(\frac{-2}{3}\right)\!\left(\frac{3p}{9-p}\right) = 1$ (includes previous M1) | M1 | Or vectors $\begin{pmatrix}9-p\\3p\end{pmatrix}\cdot\begin{pmatrix}3\\-2\end{pmatrix}$ |
| $p = 3$ | A1 [4] | |
---7 The point $A$ has coordinates $( p , 1 )$ and the point $B$ has coordinates $( 9,3 p + 1 )$, where $p$ is a constant.\\
(i) For the case where the distance $A B$ is 13 units, find the possible values of $p$.\\
(ii) For the case in which the line with equation $2 x + 3 y = 9$ is perpendicular to $A B$, find the value of $p$.
\hfill \mbox{\textit{CAIE P1 2015 Q7 [7]}}