| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Midpoint of line segment |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic coordinate geometry skills: distance formula, midpoint formula, and parallel line equations. All three parts are routine textbook exercises requiring direct application of standard formulas with no problem-solving or insight needed, making it 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 |
|---|---|---|
| \(\sqrt{(-1-4)^2 + (9 - {^-}3)^2}\) | M1 | Correct method to find line length using Pythagoras' theorem |
| \(= 13\) | A1 [2] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\dfrac{4 + {^-}1}{2}, \dfrac{^-3 + 9}{2}\right)\) | M1 | Correct method to find midpoint |
| \(\left(\dfrac{3}{2}, 3\right)\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of \(AB = -\dfrac{12}{5}\) | B1 | |
| \(y - 3 = -\dfrac{12}{5}(x - 1)\) | M1 | Correct equation for line, any gradient, through \((1, 3)\) |
| \(12x + 5y - 27 = 0\) | A1 | Correct equation in any form with gradient simplified |
| A1 [4] | \(12x + 5y - 27 = 0\) |
## Question 9:
### Part (i):
$\sqrt{(-1-4)^2 + (9 - {^-}3)^2}$ | M1 | Correct method to find line length using Pythagoras' theorem
$= 13$ | A1 [2] | cao
### Part (ii):
$\left(\dfrac{4 + {^-}1}{2}, \dfrac{^-3 + 9}{2}\right)$ | M1 | Correct method to find midpoint
$\left(\dfrac{3}{2}, 3\right)$ | A1 [2] |
### Part (iii):
Gradient of $AB = -\dfrac{12}{5}$ | B1 |
$y - 3 = -\dfrac{12}{5}(x - 1)$ | M1 | Correct equation for line, any gradient, through $(1, 3)$
$12x + 5y - 27 = 0$ | A1 | Correct equation in any form with gradient simplified
| A1 [4] | $12x + 5y - 27 = 0$
---(i) Calculate the length of $A B$.\\
(ii) Find the coordinates of the mid-point of $A B$.\\
(iii) Find the equation of the line through $( 1,3 )$ which is parallel to $A B$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\hfill \mbox{\textit{OCR C1 2009 Q9 [8]}}