| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| 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 coordinate geometry question requiring routine application of distance formula, midpoint formula, gradient calculation, and perpendicular gradient. While it involves multiple steps, each is straightforward recall with no problem-solving insight needed, making it slightly 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(AB=\sqrt{(5-2)^2+(-3-1)^2}\) | M1 | Attempt to use Pythagoras' theorem – 3/4 numbers substituted correctly and attempt to square root |
| \(AB=5\) | A1 | Final answer correct, must be fully processed. \(\pm5\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\frac{2+5}{2}, \frac{1+-3}{2}\right)\) | M1 | Correct method to find mid-point of line |
| \((3.5, -1)\) | A1 | |
| Gradient of \(AB = -\frac{4}{3}\) | B1 | Processed |
| Perpendicular gradient \(= \frac{3}{4}\) | B1ft | \(\frac{-1}{\text{their gradient}}\) processed. Alternative: M2 States P\((x,y)\) a point on the perpendicular and attempts \(PA=PB\) or \(PA^2=PB^2\); A1 At least one of PA, PB correct; A1 Both correct; M1 Expands and simplifies |
| \(y+1=\frac{3}{4}(x-\frac{7}{2})\) | M1 | Equation of straight line through their mid-point, any non-zero gradient in any form |
| A1 | ||
| \(6x-8y-29=0\) | A1 | cao Must be correct equation in required form i.e. \(k(6x-8y-29)=0\) for integer \(k\). Must have "=0" |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $AB=\sqrt{(5-2)^2+(-3-1)^2}$ | M1 | Attempt to use Pythagoras' theorem – 3/4 numbers substituted correctly **and attempt to square root** |
| $AB=5$ | A1 | Final answer correct, must be fully processed. $\pm5$ is A0 |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{2+5}{2}, \frac{1+-3}{2}\right)$ | M1 | Correct method to find mid-point of line |
| $(3.5, -1)$ | A1 | |
| Gradient of $AB = -\frac{4}{3}$ | B1 | Processed |
| Perpendicular gradient $= \frac{3}{4}$ | B1ft | $\frac{-1}{\text{their gradient}}$ processed. **Alternative:** M2 States P$(x,y)$ a point on the perpendicular and attempts $PA=PB$ or $PA^2=PB^2$; A1 At least one of PA, PB correct; A1 Both correct; M1 Expands and simplifies |
| $y+1=\frac{3}{4}(x-\frac{7}{2})$ | M1 | Equation of straight line through their mid-point, any non-zero gradient in any form |
| | A1 | |
| $6x-8y-29=0$ | A1 | **cao** Must be correct equation in required form i.e. $k(6x-8y-29)=0$ for integer $k$. **Must have "=0"** |
---5 The points $A$ and $B$ have coordinates $( 2,1 )$ and $( 5 , - 3 )$ respectively.\\
(i) Find the length of $A B$.\\
(ii) Find an equation of the line through the mid-point of $A B$ which is perpendicular 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 2015 Q5 [9]}}