| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Coordinates from geometric constraints |
| Difficulty | Moderate -0.3 This is a straightforward coordinate geometry problem requiring finding the gradient of L1, writing the equation of parallel line L2, then finding the perpendicular from A to L2. All techniques are standard (gradient, perpendicular lines, simultaneous equations, distance formula) with no novel insight needed, making it slightly easier than average but still requiring multiple steps. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Sim Eqns \(\to C(3.6, 1.8)\) | B1 B1∨ M1 M1 A1 [5] | co \(\sqrt{on gradient of \)L_1\(}. Use of \)m_1m_2 = -1$. Correct form of line eqn. co |
| (ii) \(d^2 = 1.6^2 + 3.2^2 \to d = 3.58\) | M1 A1 [2] | Correct method for \(AC\). co (accept with \(\sqrt{5}\) in answer) |
**(i)** $(2, 5)$ to $(10, 9)$ gradient $= \frac{1}{2}$
Equation of $L_1$: $y = \frac{1}{2}x$
Gradient of perpendicular $= -2$
Eqn of Perp: $y - 5 = -2(x - 2)$
Sim Eqns $\to C(3.6, 1.8)$ | B1 B1∨ M1 M1 A1 [5] | co $\sqrt{on gradient of $L_1$}. Use of $m_1m_2 = -1$. Correct form of line eqn. co
**(ii)** $d^2 = 1.6^2 + 3.2^2 \to d = 3.58$ | M1 A1 [2] | Correct method for $AC$. co (accept with $\sqrt{5}$ in answer)The line $L_1$ passes through the points $A(2, 5)$ and $B(10, 9)$. The line $L_2$ is parallel to $L_1$ and passes through the origin. The point $C$ lies on $L_2$ such that $AC$ is perpendicular to $L_2$. Find
\begin{enumerate}[label=(\roman*)]
\item the coordinates of $C$, [5]
\item the distance $AC$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2011 Q7 [7]}}