| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular from point to line |
| Difficulty | Standard +0.3 This is a standard coordinate geometry problem requiring finding the perpendicular from a point to a line. Students need to find the gradient of AB, use the perpendicular gradient property, write the equation of CX, solve simultaneously to find X, then calculate a distance ratio. While it involves multiple steps (4-5 marks typical), each step uses routine techniques taught in P1 with no novel insight required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(m\) of \(AB = -\frac{2}{3}\) | B1 | |
| \(m\) of perpendicular \(= \frac{3}{2}\) | M1 | For use of \(m_1m_2 = -1\) |
| eqn of \(AB\): \(y - 14 = -\frac{2}{3}(x-2)\) | M1 | Allow M1 for unsimplified eqn |
| eqn of \(CX\): \(y - 2 = \frac{3}{2}(x-7)\) | M1 | Allow M1 for unsimplified eqn |
| Sim Eqns \(\rightarrow X(11, 8)\) | M1 A1 [6] | For solution of sim eqns |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(AX : XB = 14-8 : 8-6 = 3:1\) or \(\sqrt{(9^2+6^2)} : \sqrt{(3^2+2^2)} = 3:1\) | M1 A1 [2] | Vector steps or Pythagoras |
## Question 7:
$A(2, 14)$, $B(14, 6)$ and $C(7, 2)$
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $m$ of $AB = -\frac{2}{3}$ | B1 | |
| $m$ of perpendicular $= \frac{3}{2}$ | M1 | For use of $m_1m_2 = -1$ |
| eqn of $AB$: $y - 14 = -\frac{2}{3}(x-2)$ | M1 | Allow M1 for unsimplified eqn |
| eqn of $CX$: $y - 2 = \frac{3}{2}(x-7)$ | M1 | Allow M1 for unsimplified eqn |
| Sim Eqns $\rightarrow X(11, 8)$ | M1 A1 [6] | For solution of sim eqns |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $AX : XB = 14-8 : 8-6 = 3:1$ or $\sqrt{(9^2+6^2)} : \sqrt{(3^2+2^2)} = 3:1$ | M1 A1 [2] | Vector steps or Pythagoras |
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\includegraphics[max width=\textwidth, alt={}, center]{13cfb59a-7781-4786-a625-919b01a2a4f0-3_465_554_255_794}
The diagram shows three points $A ( 2,14 ) , B ( 14,6 )$ and $C ( 7,2 )$. The point $X$ lies on $A B$, and $C X$ is perpendicular to $A B$. Find, by calculation,\\
(i) the coordinates of $X$,\\
(ii) the ratio $A X : X B$.
\hfill \mbox{\textit{CAIE P1 2013 Q7 [8]}}