| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Equation of line through two points |
| Difficulty | Moderate -0.3 This is a straightforward multi-part coordinate geometry question testing standard techniques: finding line equations, midpoints, distances, and perpendicularity. All parts are routine C1 exercises requiring direct application of formulas with no problem-solving insight needed. Slightly easier than average due to simple arithmetic and clear structure, but not trivial enough for -1.0 since it requires multiple techniques and careful algebraic manipulation. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \[\text{Gradient of } AB = \frac{-2-1}{-5-3} = \frac{-3}{-8} = \frac{3}{8}\] | B1 | \(\frac{3}{8}\) oe |
| \[y - 1 = \frac{3}{8}(x - 3)\] | M1 | Equation of line through either A or B, any non-zero numerical gradient |
| \[8y - 8 = 3x - 9\] | ||
| \[3x - 8y - 1 = 0\] | A1 3 | Correct equation in correct form |
| Answer | Marks | Guidance |
|---|---|---|
| \[\left(\frac{-5+3}{2}, \frac{-2+1}{2}\right)\] | M1 | Uses \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\) |
| \[= \left(-1, -\frac{1}{2}\right)\] | A1 2 | \(\left(-1, -\frac{1}{2}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \[AC = \sqrt{(-5+3)^2 + (-2-4)^2}\] | M1 | Uses \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\) |
| \[= \sqrt{2^2 + 6^2}\] | ||
| \[= \sqrt{40}\] | A1 | |
| \[= 2\sqrt{10}\] | A1 3 | Correctly simplified surd |
| Answer | Marks | Guidance |
|---|---|---|
| \[\text{Gradient of } AC = \frac{-2-4}{-5+3} = 3\] | B1 | 3 oe |
| \[\text{Gradient of } BC = \frac{4-1}{-3-3} = -\frac{1}{2}\] | B1 | \(-\frac{1}{2}\) oe |
| \[3 \times -\frac{1}{2} \neq -1\] so lines are not perpendicular | M1 | Attempts to check \(m_1 \times m_2\) |
| A1 4 | Correct conclusion www |
## 9(i)
$$\text{Gradient of } AB = \frac{-2-1}{-5-3} = \frac{-3}{-8} = \frac{3}{8}$$ | B1 | $\frac{3}{8}$ oe |
$$y - 1 = \frac{3}{8}(x - 3)$$ | M1 | Equation of line through either A or B, any non-zero numerical gradient |
$$8y - 8 = 3x - 9$$ | | |
$$3x - 8y - 1 = 0$$ | A1 3 | Correct equation in correct form |
## 9(ii)
$$\left(\frac{-5+3}{2}, \frac{-2+1}{2}\right)$$ | M1 | Uses $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ |
$$= \left(-1, -\frac{1}{2}\right)$$ | A1 2 | $\left(-1, -\frac{1}{2}\right)$ |
## 9(iii)
$$AC = \sqrt{(-5+3)^2 + (-2-4)^2}$$ | M1 | Uses $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ |
$$= \sqrt{2^2 + 6^2}$$ | | |
$$= \sqrt{40}$$ | A1 | |
$$= 2\sqrt{10}$$ | A1 3 | Correctly simplified surd |
## 9(iv)
$$\text{Gradient of } AC = \frac{-2-4}{-5+3} = 3$$ | B1 | 3 oe |
$$\text{Gradient of } BC = \frac{4-1}{-3-3} = -\frac{1}{2}$$ | B1 | $-\frac{1}{2}$ oe |
$$3 \times -\frac{1}{2} \neq -1$$ so lines are not perpendicular | M1 | Attempts to check $m_1 \times m_2$ |
| A1 4 | Correct conclusion **www** |
---9 The points $A$ and $B$ have coordinates $( - 5 , - 2 )$ and $( 3,1 )$ respectively.\\
(i) Find the equation of the line $A B$, giving your answer in the form $a x + b y + c = 0$.\\
(ii) Find the coordinates of the mid-point of $A B$.
The point $C$ has coordinates (-3,4).\\
(iii) Calculate the length of $A C$, giving your answer in simplified surd form.\\
(iv) Determine whether the line $A C$ is perpendicular to the line $B C$, showing all your working.
\hfill \mbox{\textit{OCR C1 2008 Q9 [12]}}