| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular bisector of segment |
| Difficulty | Moderate -0.8 This is a straightforward multi-part coordinate geometry question testing standard techniques: midpoint formula, perpendicular gradient, point-line verification, and area calculation. All steps are routine applications of formulas with no problem-solving insight required, making it easier than average but not trivial due to the multiple parts. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.10g Problem solving with vectors: in geometry |
| Answer | Marks |
|---|---|
| Midpoint \(=\left(\frac{1+3}{2},\frac{1+5}{2}\right)=(2,3)\) | B1 |
| Grad of AC \(\frac{5-1}{3-1}=2 \Rightarrow\) Grad of perp \(=-\frac{1}{2}\) | M1 A1 |
| \(\Rightarrow y-3=-\frac{1}{2}(x-2) \Rightarrow -2y+6=x-2 \Rightarrow x+2y=8\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(-2+10=8\), so B lies on the line | B1 |
| B to \((2,3)=\binom{4}{-2} \Rightarrow (2,3)\) to \(D=\binom{4}{-2} \Rightarrow D\) is \((6,1)\) | M1 A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(MA=\sqrt{(2-1)^2+(3-1)^2}=\sqrt{5}\) | M1 | Use to find area (Both) |
| \(MB=\sqrt{(2-{-2})^2+(3-5)^2}=\sqrt{20}\) | A1 | |
| Area \(= 2 \cdot MA \cdot MB = 2\sqrt{5}\cdot\sqrt{20}=20\) | A1 |
## Question 10(i):
Midpoint $=\left(\frac{1+3}{2},\frac{1+5}{2}\right)=(2,3)$ | B1 |
Grad of AC $\frac{5-1}{3-1}=2 \Rightarrow$ Grad of perp $=-\frac{1}{2}$ | M1 A1 |
$\Rightarrow y-3=-\frac{1}{2}(x-2) \Rightarrow -2y+6=x-2 \Rightarrow x+2y=8$ | M1 A1 |
**Total: 5**
## Question 10(ii):
$-2+10=8$, so B lies on the line | B1 |
B to $(2,3)=\binom{4}{-2} \Rightarrow (2,3)$ to $D=\binom{4}{-2} \Rightarrow D$ is $(6,1)$ | M1 A1 A1 |
**Total: 4**
## Question 10(iii):
$MA=\sqrt{(2-1)^2+(3-1)^2}=\sqrt{5}$ | M1 | Use to find area (Both)
$MB=\sqrt{(2-{-2})^2+(3-5)^2}=\sqrt{20}$ | A1 |
Area $= 2 \cdot MA \cdot MB = 2\sqrt{5}\cdot\sqrt{20}=20$ | A1 |
**Total: 3**
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b927f8b-ddf8-481d-a1ce-3b90bb1435f0-3_437_572_1058_538}
\captionsetup{labelformat=empty}
\caption{Fig. 10}
\end{center}
\end{figure}
In Fig.10, A has coordinates $( 1,1 )$ and C has coordinates $( 3,5 )$. M is the mid-point of AC . The line $l$ is perpendicular to AC.\\
(i) Find the coordinates of M .
Hence find the equation of $l$.\\
(ii) The point B has coordinates $( - 2,5 )$.
Show that B lies on the line $l$.\\
Find the coordinates of the point D such that ABCD is a rhombus.\\
(iii) Find the lengths MC and MB .
Hence calculate the area of the rhombus ABCD .
\hfill \mbox{\textit{OCR MEI C1 Q10 [12]}}