| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Verify shape type from coordinates |
| Difficulty | Standard +0.3 This is a multi-part coordinate geometry question requiring standard techniques (gradient for perpendicularity, distance formula, line equations, midpoint formula) applied systematically across 6 parts. While lengthy with multiple steps, each individual part uses routine C1 methods without requiring novel insight or complex problem-solving, making it slightly easier than average overall. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Grad \(AB = \frac{4-3}{2--2} = \frac{1}{4}\) | B1 | Both |
| Answer | Marks |
|---|---|
| \(\Rightarrow m_1m_2=-1 \Rightarrow\) perpendicular | B1 |
| Answer | Marks |
|---|---|
| \(AB = \sqrt{(4-3)^2+(2+2)^2} = \sqrt{17}\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow\) Area \(= \sqrt{17}\cdot\sqrt{68} = \sqrt{17}\cdot2\sqrt{17} = 34\) | M1 A1 | c.a.o |
| Answer | Marks |
|---|---|
| \(CD: \frac{y+4}{-3+4} = \frac{x-4}{8-4} \Rightarrow 4y+16=x-4\) | M1 |
| Answer | Marks |
|---|---|
| \(\Rightarrow x=0,\ y=-5\), i.e. \(X(0,-5)\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Midpoint of \(BX\) is \(\left(\frac{2+0}{2}, \frac{4-5}{2}\right) = \left(1,-\frac{1}{2}\right)\) | M1 | Attempt at any valid method |
| Midpoint of \(AD\) is \(\left(\frac{-2+4}{2}, \frac{3-4}{2}\right) = \left(1,-\frac{1}{2}\right)\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Parallelogram \(=\) Trapezium \(MBCD +\) Triangle \(MAB\); Triangle \(BXC =\) Trapezium \(MBCD +\) Triangle \(MXD\); Triangle \(MAB \cong\) Triangle \(MXD\) (SAS), so area of triangle \(BXC = 34\) | B1 | |
| \(BX = \sqrt{(2-0)^2+(4+5)^2} = \sqrt{85}\) | M1 | |
| \(\Rightarrow\) Area \(= \frac{1}{2}BX \times \text{Perp} = 34 \Rightarrow \text{Perp} = \frac{68}{\sqrt{85}} = \frac{4}{5}\sqrt{85}\) | A1 | Or equivalent |
## Question 12:
**(i)**
Grad $AB = \frac{4-3}{2--2} = \frac{1}{4}$ | B1 | Both
Grad $BD = \frac{-4-4}{4-2} = -4$
$\Rightarrow m_1m_2=-1 \Rightarrow$ perpendicular | B1 |
**(ii)**
$AB = \sqrt{(4-3)^2+(2+2)^2} = \sqrt{17}$ | M1 |
$BD = \sqrt{(-4-4)^2+(4-2)^2} = \sqrt{68}$
$\Rightarrow$ Area $= \sqrt{17}\cdot\sqrt{68} = \sqrt{17}\cdot2\sqrt{17} = 34$ | M1 A1 | c.a.o
**(iii)**
$CD: \frac{y+4}{-3+4} = \frac{x-4}{8-4} \Rightarrow 4y+16=x-4$ | M1 |
$\Rightarrow 4y=x-20$
$\Rightarrow x=0,\ y=-5$, i.e. $X(0,-5)$ | E1 |
**(iv)**
Midpoint of $BX$ is $\left(\frac{2+0}{2}, \frac{4-5}{2}\right) = \left(1,-\frac{1}{2}\right)$ | M1 | Attempt at any valid method
Midpoint of $AD$ is $\left(\frac{-2+4}{2}, \frac{3-4}{2}\right) = \left(1,-\frac{1}{2}\right)$ | E1 |
**(v)**
Parallelogram $=$ Trapezium $MBCD +$ Triangle $MAB$; Triangle $BXC =$ Trapezium $MBCD +$ Triangle $MXD$; Triangle $MAB \cong$ Triangle $MXD$ (SAS), so area of triangle $BXC = 34$ | B1 |
$BX = \sqrt{(2-0)^2+(4+5)^2} = \sqrt{85}$ | M1 |
$\Rightarrow$ Area $= \frac{1}{2}BX \times \text{Perp} = 34 \Rightarrow \text{Perp} = \frac{68}{\sqrt{85}} = \frac{4}{5}\sqrt{85}$ | A1 | Or equivalent12 ABCD is a parallelogram. The coordinates of $\mathrm { A } , \mathrm { B } , \mathrm { C }$ and D are (-2, 3), (2, 4), (8, -3) and ( $4 , - 4$ ) respectively.\\
\includegraphics[max width=\textwidth, alt={}, center]{3b6291ef-bef9-49de-a20f-591e621bed65-4_592_725_387_492}\\
(i) Prove that AB and BD are perpendicular.\\
(ii) Find the lengths of AB and BD and hence find the area of the parallelogram ABCD\\
(iii) Find the equation of the line CD and show that it meets the $y$-axis at $\mathrm { X } ( 0 , - 5 )$.\\
(iv) Show that the lines BX and AD bisect each other.\\
(v) Explain why the area of the parallelogram ABCD is equal to the area of the triangle BXC.\\
Find the length of BX and hence calculate exactly the perpendicular distance of C from BX .
\hfill \mbox{\textit{OCR MEI C1 Q12 [12]}}