| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Moderate -0.8 This is a straightforward C1 coordinate geometry question requiring direct application of standard formulas: midpoint formula and perpendicular gradient. Part (i) is pure recall, part (ii) involves finding gradient of AB, taking negative reciprocal, then using point-slope form—all routine procedures with no problem-solving insight needed. Easier than average A-level questions. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| i) \(\left(\frac{5 + (-1) + 7 + (-5)}{2}, \frac{2}{2}\right)\); \((2, 1)\) | M1, A1 [2] | Correct method to find midpoint of line. At least 3 out of 4 terms correctly substituted |
| ii) Gradient of AB = \(\frac{7 - (-5)}{5 - (-1)} = 2\); Perpendicular gradient = \(-\frac{1}{2}\); \(y - 7 = -\frac{1}{2}(x - 5)\); \(x + 2y - 19 = 0\) | B1, B1n, M1, A1n, A1 [5] | Gradient of AB correctly found as 2. Fully processed \(\frac{-1}{\text{their gradient}}\). Equation of straight line through A or B, any non-zero gradient. Equation of straight line through A only, their perpendicular gradient, in any form. Correct equation in given form. i.e. \(k(x + 2y - 19) = 0\) for integer \(k\). Must have "=0". |
**i)** $\left(\frac{5 + (-1) + 7 + (-5)}{2}, \frac{2}{2}\right)$; $(2, 1)$ | M1, A1 [2] | Correct method to find midpoint of line. At least 3 out of 4 terms correctly substituted
**ii)** Gradient of AB = $\frac{7 - (-5)}{5 - (-1)} = 2$; Perpendicular gradient = $-\frac{1}{2}$; $y - 7 = -\frac{1}{2}(x - 5)$; $x + 2y - 19 = 0$ | B1, B1n, M1, A1n, A1 [5] | Gradient of AB correctly found as 2. Fully processed $\frac{-1}{\text{their gradient}}$. Equation of straight line through A or B, any non-zero gradient. Equation of straight line through A only, their perpendicular gradient, in any form. Correct equation in given form. i.e. $k(x + 2y - 19) = 0$ for integer $k$. Must have "=0".$A$ is the point $(5, 7)$ and $B$ is the point $(-1, -5)$.
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of the mid-point of the line segment $AB$. [2]
\item Find an equation of the line through $A$ that is perpendicular to the line segment $AB$, giving your answer in the form $ax + by + c = 0$ where $a$, $b$ and $c$ are integers. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 2014 Q7 [7]}}