| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Coordinates from geometric constraints |
| Difficulty | Standard +0.3 This is a multi-part coordinate geometry question requiring gradient calculation, simultaneous equations to find point C, and perpendicular bisector work. While it has several steps (typical of 3-part questions worth ~10 marks), each individual technique is standard P1 material: gradient formula, equation of a line, and finding intersections. The constraint that gradients are related by factor m adds mild problem-solving, but the question guides students through each step methodically without requiring novel insight. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| \(y\)-step \(\div\) \(x\)-step \(= 2\), \(\rightarrow m = 1\) | M1 A1 [2] | Gradient \(=\) \(y\)-step \(\div\) \(x\)-step used. co |
| Answer | Marks | Guidance |
|---|---|---|
| Sim eqns: \(y + 2x = 4\), \(y = x + 7\); \(\rightarrow C(-1,\ 6)\) | M1 A1\(\sqrt{}\), A1\(\sqrt{}\), A1 [4] | Correct form of one of lines. \(\sqrt{}\) to his \(m\); \(\sqrt{}\) to his \(m\); co |
| Answer | Marks | Guidance |
|---|---|---|
| \(M\) is \((9, 10)\) | B1 | co |
| Perp gradient is \(-\frac{1}{2}\); \(\rightarrow 2y + x = 29\), \(y = x + 7\); Sim eqns \(\rightarrow D(5,\ 12)\) | M1, M1, A1 [4] | Use of \(m_1 m_2 = -1\); Solve sim eqns for their \(BC\) & perp. bis; co |
## Question 8:
**Part (i):**
$y$-step $\div$ $x$-step $= 2$, $\rightarrow m = 1$ | M1 A1 [2] | Gradient $=$ $y$-step $\div$ $x$-step used. co
**Part (ii):**
Eqn of $AC$: $y + 2 = -2(x-3)$; Eqn of $BC$: $y - 22 = (x-15)$
Sim eqns: $y + 2x = 4$, $y = x + 7$; $\rightarrow C(-1,\ 6)$ | M1 A1$\sqrt{}$, A1$\sqrt{}$, A1 [4] | Correct form of one of lines. $\sqrt{}$ to his $m$; $\sqrt{}$ to his $m$; co
**Part (iii):**
$M$ is $(9, 10)$ | B1 | co
Perp gradient is $-\frac{1}{2}$; $\rightarrow 2y + x = 29$, $y = x + 7$; Sim eqns $\rightarrow D(5,\ 12)$ | M1, M1, A1 [4] | Use of $m_1 m_2 = -1$; Solve sim eqns for their $BC$ & perp. bis; co
---8\\
\includegraphics[max width=\textwidth, alt={}, center]{56d4d40a-32f5-4f2d-938e-a24312cd42e7-3_625_547_1489_797}
The diagram shows a triangle $A B C$ in which $A$ is $( 3 , - 2 )$ and $B$ is $( 15,22 )$. The gradients of $A B , A C$ and $B C$ are $2 m , - 2 m$ and $m$ respectively, where $m$ is a positive constant.\\
(i) Find the gradient of $A B$ and deduce the value of $m$.\\
(ii) Find the coordinates of $C$.
The perpendicular bisector of $A B$ meets $B C$ at $D$.\\
(iii) Find the coordinates of $D$.
\hfill \mbox{\textit{CAIE P1 2010 Q8 [10]}}