| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Verify collinearity or parallel/perpendicular relationship |
| Difficulty | Standard +0.3 This is a straightforward coordinate geometry question requiring standard techniques: calculating gradients to verify parallel lines, using distance formula to check if sides are equal, finding intersection of two lines, and verifying midpoints. All parts are routine applications of basic formulas with no novel problem-solving 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 relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| grad \(CD = \dfrac{5-3}{3-(-1)} = \dfrac{2}{4}\) o.e. isw | M1 | NB needs to be obtained independently of grad AB |
| grad \(AB = \dfrac{3-(-1)}{6-(-2)}\) or \(\dfrac{4}{8}\) isw | M1 | |
| same gradient so parallel www | A1 | must be explicit conclusion mentioning 'same gradient' or 'parallel'; if M0, allow B1 for 'parallel lines have same gradient' o.e. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([BC^2 =] 3^2 + 2^2\) | M1 | accept \((6-3)^2 + (3-5)^2\) o.e. |
| \([BC^2 =] 13\) | A1 | or \([BC =] \sqrt{13}\) |
| showing \(AD^2 = 1^2 + 4^2 [= 17] [\neq BC^2]\) isw | A1 | or \([AD =] \sqrt{17}\); or equivalent marks for finding AD or \(AD^2\) first; alt method: showing \(AC \neq BD\) — mark equivalently |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| [BD equation is] \(y = 3\) | M1 | e.g. allow for 'at M, \(y = 3\)' or for 3 substituted in equation of AC |
| equation of AC is \(y - 5 = \frac{6}{5}(x-3)\) o.e. \([y = 1.2x + 1.4\) o.e.] | M2 | or M1 for grad \(AC = \frac{6}{5}\) o.e. (accept unsimplified) and M1 for using their grad of AC with coords of \(A(-2,-1)\) or \(C(3,5)\) in equation of line or M1 for 'stepping' method to reach M |
| M is \(\left(\frac{4}{3}, 3\right)\) o.e. isw | A1 | allow: at M, \(x = \frac{16}{12}\) o.e. [e.g. \(= \frac{4}{3}\)] isw; A0 for 1.3 without a fraction answer seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| midpoint of \(BD = \left(\frac{5}{2}, 3\right)\) or equivalent simplified form cao | M1 | or showing \(BM \neq MD\) o.e. \([BM = \frac{14}{3},\ MD = \frac{7}{3}]\) |
| midpoint \(AC = \left(\frac{1}{2}, 2\right)\) or equivalent simplified form cao or 'M is \(\frac{2}{3}\) of way from A to C' | M1 | or showing \(AM \neq MC\) or \(AM^2 \neq MC^2\) |
| conclusion 'neither diagonal bisects the other' | A1 | in these methods A1 is dependent on coords of M having been obtained in part (iii) or in this part; coordinates of M need not be correct; it is also dependent on midpoints of both AC and BD attempted, at least one correct; alt method: show that midpoint of BD does not lie on AC (M1) and vice-versa (M1), A1 for both and conclusion |
## Question 4(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| grad $CD = \dfrac{5-3}{3-(-1)} = \dfrac{2}{4}$ o.e. isw | M1 | NB needs to be obtained independently of grad AB |
| grad $AB = \dfrac{3-(-1)}{6-(-2)}$ or $\dfrac{4}{8}$ isw | M1 | |
| same gradient so parallel www | A1 | must be explicit conclusion mentioning 'same gradient' or 'parallel'; if M0, allow B1 for 'parallel lines have same gradient' o.e. |
---
## Question 4(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[BC^2 =] 3^2 + 2^2$ | M1 | accept $(6-3)^2 + (3-5)^2$ o.e. |
| $[BC^2 =] 13$ | A1 | or $[BC =] \sqrt{13}$ |
| showing $AD^2 = 1^2 + 4^2 [= 17] [\neq BC^2]$ isw | A1 | or $[AD =] \sqrt{17}$; or equivalent marks for finding AD or $AD^2$ first; alt method: showing $AC \neq BD$ — mark equivalently |
---
## Question 4(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| [BD equation is] $y = 3$ | M1 | e.g. allow for 'at M, $y = 3$' or for 3 substituted in equation of AC |
| equation of AC is $y - 5 = \frac{6}{5}(x-3)$ o.e. $[y = 1.2x + 1.4$ o.e.] | M2 | or M1 for grad $AC = \frac{6}{5}$ o.e. (accept unsimplified) and M1 for using their grad of AC with coords of $A(-2,-1)$ or $C(3,5)$ in equation of line or M1 for 'stepping' method to reach M |
| M is $\left(\frac{4}{3}, 3\right)$ o.e. isw | A1 | allow: at M, $x = \frac{16}{12}$ o.e. [e.g. $= \frac{4}{3}$] isw; A0 for 1.3 without a fraction answer seen |
---
## Question 4(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| midpoint of $BD = \left(\frac{5}{2}, 3\right)$ or equivalent simplified form cao | M1 | or showing $BM \neq MD$ o.e. $[BM = \frac{14}{3},\ MD = \frac{7}{3}]$ |
| midpoint $AC = \left(\frac{1}{2}, 2\right)$ or equivalent simplified form cao or 'M is $\frac{2}{3}$ of way from A to C' | M1 | or showing $AM \neq MC$ or $AM^2 \neq MC^2$ |
| conclusion 'neither diagonal bisects the other' | A1 | in these methods A1 is dependent on coords of M having been obtained in part (iii) or in this part; coordinates of M need not be correct; it is also dependent on midpoints of both AC and BD attempted, at least one correct; alt method: show that midpoint of BD does not lie on AC (M1) and vice-versa (M1), A1 for both and conclusion |4
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{01569a16-66ba-422e-a74d-6f9430dd245b-2_592_782_322_730}
\captionsetup{labelformat=empty}
\caption{Fig. 10}
\end{center}
\end{figure}
Fig. 10 shows a trapezium ABCD . The coordinates of its vertices are $\mathrm { A } ( - 2 , - 1 ) , \mathrm { B } ( 6,3 ) , \mathrm { C } ( 3,5 )$ and $\mathrm { D } ( - 1,3 )$.\\
(i) Verify that the lines AB and DC are parallel.\\
(ii) Prove that the trapezium is not isosceles.\\
(iii) The diagonals of the trapezium meet at M . Find the exact coordinates of M .\\
(iv) Show that neither diagonal of the trapezium bisects the other.
\hfill \mbox{\textit{OCR MEI C1 Q4 [13]}}