| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Perpendicular bisector of chord |
| Difficulty | Easy -1.2 This is a straightforward two-part question testing basic coordinate geometry. Part (a) requires finding the midpoint and negative reciprocal gradient (standard procedure), while part (b) uses the distance formula for the radius. Both are routine textbook exercises with no problem-solving or insight required, making this easier than average. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| Mid-point \(AB\) is \(\left(\dfrac{10+5}{2}, \dfrac{2-1}{2}\right) = \left(\dfrac{15}{2}, \dfrac{1}{2}\right)\) | B1 | Accept unsimplified |
| Gradient of \(AB = \dfrac{-1-2}{10-5} = \dfrac{-3}{5}\), Gradient perpendicular \(= \dfrac{5}{3}\) | M1 | For use of \(\dfrac{\text{Change in }y}{\text{Change in }x}\), condone inconsistent order of \(x\) and \(y\), and \(m_1 m_2 = -1\) |
| \(\dfrac{y - \frac{1}{2}}{x - \frac{15}{2}} = \dfrac{5}{3}\) \(\left[y - \dfrac{1}{2} = \dfrac{5}{3}\left(x - \dfrac{15}{2}\right)\right]\) | A1 | OE ISW. Any correct version e.g. \(y = \dfrac{5}{3}x - 12\) or \(5x - 3y = 36\) |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\text{Radius} =] \sqrt{34}\) or \(5.8\) AWRT or \([(\text{radius})^2 =] 34\) | B1 | Sight of \(\sqrt{34}\) or \(34\). Condone confusion of \(r\) and \(r^2\) |
| \((x-5)^2 + (y-2)^2\) | B1 | Sight of \((x-5)^2 + (y-2)^2\) |
| \((x-5)^2 + (y-2)^2 = 34\) | B1 | CAO ISW |
| Alternative method: | ||
| \(x^2 + y^2 - 10x - 4y\) | B1 | |
| \([c=]\ 5\) or \([c=]\ {-5}\) | B1 | Substitution of \((10,-1)\) into \(x^2 + y^2 - 10x - 4y + c = 0\) |
| \(x^2 + y^2 - 10x - 4y - 5 = 0\) | B1 | |
| Total | 3 |
## Question 1(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mid-point $AB$ is $\left(\dfrac{10+5}{2}, \dfrac{2-1}{2}\right) = \left(\dfrac{15}{2}, \dfrac{1}{2}\right)$ | B1 | Accept unsimplified |
| Gradient of $AB = \dfrac{-1-2}{10-5} = \dfrac{-3}{5}$, Gradient perpendicular $= \dfrac{5}{3}$ | M1 | For use of $\dfrac{\text{Change in }y}{\text{Change in }x}$, condone inconsistent order of $x$ and $y$, and $m_1 m_2 = -1$ |
| $\dfrac{y - \frac{1}{2}}{x - \frac{15}{2}} = \dfrac{5}{3}$ $\left[y - \dfrac{1}{2} = \dfrac{5}{3}\left(x - \dfrac{15}{2}\right)\right]$ | A1 | OE ISW. Any correct version e.g. $y = \dfrac{5}{3}x - 12$ or $5x - 3y = 36$ |
| **Total** | **3** | |
## Question 1(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\text{Radius} =] \sqrt{34}$ or $5.8$ AWRT or $[(\text{radius})^2 =] 34$ | B1 | Sight of $\sqrt{34}$ or $34$. Condone confusion of $r$ and $r^2$ |
| $(x-5)^2 + (y-2)^2$ | B1 | Sight of $(x-5)^2 + (y-2)^2$ |
| $(x-5)^2 + (y-2)^2 = 34$ | B1 | CAO ISW |
| **Alternative method:** | | |
| $x^2 + y^2 - 10x - 4y$ | B1 | |
| $[c=]\ 5$ or $[c=]\ {-5}$ | B1 | Substitution of $(10,-1)$ into $x^2 + y^2 - 10x - 4y + c = 0$ |
| $x^2 + y^2 - 10x - 4y - 5 = 0$ | B1 | |
| **Total** | **3** | |1 Points $A$ and $B$ have coordinates $( 5,2 )$ and $( 10 , - 1 )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the perpendicular bisector of $A B$.
\item Find the equation of the circle with centre $A$ which passes through $B$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q1 [6]}}