| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular bisector of segment |
| Difficulty | Moderate -0.3 This is a standard C1 coordinate geometry question testing routine skills: parallel lines (same gradient), distance formula, and perpendicular bisector. All three parts use well-practiced techniques with no problem-solving insight required, making it slightly easier than average but not trivial due to the multi-step nature and surd simplification. |
| 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 | Marks | Guidance |
| Gradient \(= 4\) | B1 | Gradient of 4 soi |
| \(y - 7 = 4(x-2)\) | M1 | Attempts equation of straight line through \((2,7)\) with any gradient |
| \(y = 4x - 1\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\) | M1 | Use of correct formula for \(d\) or \(d^2\) (3 values correctly substituted) |
| \(= \sqrt{(2-{-1})^2+(7-{-2})^2}\) | ||
| \(= \sqrt{3^2+9^2}\) | A1 | \(\sqrt{3^2+9^2}\) |
| \(= \sqrt{90} = 3\sqrt{10}\) | A1 [3] | Correct simplified surd |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient of \(AB = 3\) | B1 | |
| Gradient of perpendicular line \(= -\frac{1}{3}\) | B1 ft | SR Allow B1 for \(-\frac{1}{4}\) |
| Midpoint of \(AB = \left(\frac{1}{2}, \frac{5}{2}\right)\) | B1 | |
| \(y - \frac{5}{2} = -\frac{1}{3}\left(x - \frac{1}{2}\right)\) | M1 | Attempts equation of straight line through their midpoint with any non-zero gradient |
| \(x + 3y - 8 = 0\) | A1, A1 [6+3+3=12] | \(y - \frac{5}{2} = \frac{-1}{3}\left(x - \frac{1}{2}\right)\); \(x+3y-8=0\) |
## Question 9:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient $= 4$ | B1 | Gradient of 4 soi |
| $y - 7 = 4(x-2)$ | M1 | Attempts equation of straight line through $(2,7)$ with any gradient |
| $y = 4x - 1$ | A1 [3] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ | M1 | Use of correct formula for $d$ or $d^2$ (3 values correctly substituted) |
| $= \sqrt{(2-{-1})^2+(7-{-2})^2}$ | | |
| $= \sqrt{3^2+9^2}$ | A1 | $\sqrt{3^2+9^2}$ |
| $= \sqrt{90} = 3\sqrt{10}$ | A1 [3] | Correct simplified surd |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of $AB = 3$ | B1 | |
| Gradient of perpendicular line $= -\frac{1}{3}$ | B1 ft | **SR** Allow B1 for $-\frac{1}{4}$ |
| Midpoint of $AB = \left(\frac{1}{2}, \frac{5}{2}\right)$ | B1 | |
| $y - \frac{5}{2} = -\frac{1}{3}\left(x - \frac{1}{2}\right)$ | M1 | Attempts equation of straight line through their midpoint with any non-zero gradient |
| $x + 3y - 8 = 0$ | A1, A1 [6+3+3=12] | $y - \frac{5}{2} = \frac{-1}{3}\left(x - \frac{1}{2}\right)$; $x+3y-8=0$ |
---(i) Find the equation of the line through $A$ parallel to the line $y = 4 x - 5$, giving your answer in the form $y = m x + c$.\\
(ii) Calculate the length of $A B$, giving your answer in simplified surd form.\\
(iii) Find the equation of the line which passes through the mid-point of $A B$ and which is perpendicular to $A B$. Give your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\hfill \mbox{\textit{OCR C1 2007 Q9 [12]}}