| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular bisector of segment |
| Difficulty | Moderate -0.3 This is a straightforward coordinate geometry question requiring standard techniques: finding midpoint, gradient of perpendicular line, and solving simultaneous equations (one linear, one from distance formula). Part (i) is routine bookwork, while part (ii) adds a modest problem-solving element by combining the perpendicular bisector equation with a circle equation, but the approach is clear and the algebra is manageable. Slightly easier than average due to the step-by-step guidance provided. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) mid-point \(= (3, 4)\) | B1 | soi |
| Grad. \(AB = -\frac{1}{2}\) \(\rightarrow\) grad. of perp. \(= 2\) | M1 | For use of \(-1/m\) soi ft on their (3, 4) and 2 |
| \(y - 4 = 2(x - 3)\) | M1 | |
| \(y = 2x - 2\) | A1 | [4] |
| (ii) \(q = 2p - 2\) | B1 | ft for 1st eqn. |
| \(p^2 + q^2 = 4\) oe | B1 | |
| \(p^2 + (2p-2)^2 = 4 \Rightarrow 5p^2 - 8p = 0\) | M1 | Attempt substn (linear into quadratic) & simplify |
| \(\{OR \frac{1}{4}(q+2)^2 + q^2 = 4 \Rightarrow 5q^2 + 4q - 12 = 0\}\) | ||
| \((0, -2)\) and \((\frac{8}{5}, \frac{6}{5})\) | A1A1 | [5] |
(i) mid-point $= (3, 4)$ | B1 | soi
Grad. $AB = -\frac{1}{2}$ $\rightarrow$ grad. of perp. $= 2$ | M1 | For use of $-1/m$ soi ft on their (3, 4) and 2
$y - 4 = 2(x - 3)$ | M1 |
$y = 2x - 2$ | A1 | [4]
(ii) $q = 2p - 2$ | B1 | ft for 1st eqn.
$p^2 + q^2 = 4$ oe | B1 |
$p^2 + (2p-2)^2 = 4 \Rightarrow 5p^2 - 8p = 0$ | M1 | Attempt substn (linear into quadratic) & simplify
$\{OR \frac{1}{4}(q+2)^2 + q^2 = 4 \Rightarrow 5q^2 + 4q - 12 = 0\}$ | |
$(0, -2)$ and $(\frac{8}{5}, \frac{6}{5})$ | A1A1 | [5]7 The point $A$ has coordinates ( $- 1,6$ ) and the point $B$ has coordinates (7,2).\\
(i) Find the equation of the perpendicular bisector of $A B$, giving your answer in the form $y = m x + c$.\\
(ii) A point $C$ on the perpendicular bisector has coordinates $( p , q )$. The distance $O C$ is 2 units, where $O$ is the origin. Write down two equations involving $p$ and $q$ and hence find the coordinates of the possible positions of $C$.
\hfill \mbox{\textit{CAIE P1 2013 Q7 [9]}}