| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular from point to line |
| Difficulty | Moderate -0.8 This is a straightforward C1 coordinate geometry question requiring standard techniques: finding perpendicular gradient, line equations, intersection points, distance formula, and triangle area. All parts follow routine procedures with no problem-solving insight needed, making it easier than average but not trivial due to the multi-step nature and algebraic manipulation required. |
| 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 | Marks | Guidance |
| \((0, 14)\) and \((\frac{14}{4}, 0)\) o.e. isw | 4 | M2 for evidence of correct use of gradient with \((2, 6)\), e.g. sketch with 'stepping' or \(y - 6 = -4(x - 2)\) seen or \(y = -4x + 14\) o.e. |
| M1 for \(y = -4x + c\) [accept any letter or number] and M1 for \(6 = -4 \times 2 + c\) | ||
| A1 for \((0, 14)\) [\(c = 14\) is not sufficient for A1] and A1 for \((\frac{14}{4}, 0)\) o.e.; allow when \(x = 0\), \(y = 14\) etc isw |
## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0, 14)$ and $(\frac{14}{4}, 0)$ o.e. isw | 4 | M2 for evidence of correct use of gradient with $(2, 6)$, e.g. sketch with 'stepping' or $y - 6 = -4(x - 2)$ seen or $y = -4x + 14$ o.e. |
| | | M1 for $y = -4x + c$ [accept any letter or number] and M1 for $6 = -4 \times 2 + c$ |
| | | A1 for $(0, 14)$ [$c = 14$ is not sufficient for A1] and A1 for $(\frac{14}{4}, 0)$ o.e.; allow when $x = 0$, $y = 14$ etc isw |
---5 A line has gradient - 4 and passes through the point (2,6). Find the coordinates of its points of intersection with the axes.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{01569a16-66ba-422e-a74d-6f9430dd245b-3_433_835_353_715}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
Fig. 11 shows the line joining the points $\mathrm { A } ( 0,3 )$ and $\mathrm { B } ( 6,1 )$.\\
(i) Find the equation of the line perpendicular to AB that passes through the origin, O .\\
(ii) Find the coordinates of the point where this perpendicular meets AB .\\
(iii) Show that the perpendicular distance of AB from the origin is $\frac { 9 \sqrt { 10 } } { 10 }$.\\
(iv) Find the length of AB , expressing your answer in the form $a \sqrt { 10 }$.\\
(v) Find the area of triangle OAB .
\hfill \mbox{\textit{OCR MEI C1 Q5 [4]}}