| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Moderate -0.3 This is a straightforward coordinate geometry question requiring students to find the gradient of the perpendicular line (negative reciprocal) and use point-slope form. It's slightly easier than average because it's a standard textbook exercise with clear steps: rearrange to find gradient of l, find perpendicular gradient, substitute point to find c. No problem-solving insight required. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
\begin{enumerate}
\item The straight line $l$ has the equation $x - 5 y = 7$.
\end{enumerate}
The straight line $m$ is perpendicular to $l$ and passes through the point $( - 4,1 )$.\\
Find an equation for $m$ in the form $y = m x + c$.\\
\hfill \mbox{\textit{OCR C1 Q2 [4]}}