| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Parameter from distance condition |
| Difficulty | Moderate -0.3 This is a straightforward coordinate geometry question requiring distance formula application and solving a quadratic equation in part (i), plus simple mid-point calculation in part (ii). While it involves multiple steps, all techniques are standard C1 material with no conceptual challenges—slightly easier than average due to routine 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 |
|---|---|---|
| (i) Length AC \(= \sqrt{(8-5)^2 + (2-1)^2} = \sqrt{3^2 + 1^2} = \sqrt{10}\); Length AB \(= \sqrt{(p-5)^2 + (7-1)^2} = \sqrt{(p-5)^2 + 36}\); \(\sqrt{(p-5)^2 + 36} = 2\sqrt{10}\); \(p^2 - 10p + 25 + 36 = 40\); \(p^2 - 10p + 21 = 0\); \((p-7)(p-3) = 0\); \(p = 7, 3\) | M1, A1, A1, M1, M1, A1, A1, 7 | Uses \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\); \(\sqrt{10}\) (± \(\sqrt{10}\) scores A0); \(\sqrt{(p-5)^2 + (7-1)^2}\); AB = 2AC (with algebraic expression) used; Obtains 3 term quadratic \(= 0\) suitable for solving or \((p-5)^2 = 4\)___; \(p = 7\); \(p = 3\) SR: If no working seen, and one correct value found, award B2 in place of the final 4 marks in part (i) |
| (ii) \(7 = 3x - 14\); \(x = 7\); \((5, 1)\) \((7, 7)\); Mid-point \((6, 4)\) | M1, A1, M1, A1√, 4 | Correct method to find x; \(x = 7\); Use \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\); \((6, 4)\) or correct midpoint for their AB Alternative method: y coordinate of midpoint \(= 4\) M1 A1; sub 4 into equation of line M1; obtains \(x = 6\) A1 |
**(i)** Length AC $= \sqrt{(8-5)^2 + (2-1)^2} = \sqrt{3^2 + 1^2} = \sqrt{10}$; Length AB $= \sqrt{(p-5)^2 + (7-1)^2} = \sqrt{(p-5)^2 + 36}$; $\sqrt{(p-5)^2 + 36} = 2\sqrt{10}$; $p^2 - 10p + 25 + 36 = 40$; $p^2 - 10p + 21 = 0$; $(p-7)(p-3) = 0$; $p = 7, 3$ | M1, A1, A1, M1, M1, A1, A1, 7 | Uses $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$; $\sqrt{10}$ (± $\sqrt{10}$ scores A0); $\sqrt{(p-5)^2 + (7-1)^2}$; AB = 2AC (with algebraic expression) used; Obtains 3 term quadratic $= 0$ suitable for solving or $(p-5)^2 = 4$___; $p = 7$; $p = 3$ **SR:** If no working seen, and one correct value found, award B2 in place of the final 4 marks in part (i)
**(ii)** $7 = 3x - 14$; $x = 7$; $(5, 1)$ $(7, 7)$; Mid-point $(6, 4)$ | M1, A1, M1, A1√, 4 | Correct method to find x; $x = 7$; Use $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$; $(6, 4)$ or correct midpoint for their AB **Alternative method:** y coordinate of midpoint $= 4$ M1 A1; sub 4 into equation of line M1; obtains $x = 6$ A19 The points $A , B$ and $C$ have coordinates $( 5,1 ) , ( p , 7 )$ and $( 8,2 )$ respectively.\\
(i) Given that the distance between points $A$ and $B$ is twice the distance between points $A$ and $C$, calculate the possible values of $p$.\\
(ii) Given also that the line passing through $A$ and $B$ has equation $y = 3 x - 14$, find the coordinates of the mid-point of $A B$.
\hfill \mbox{\textit{OCR C1 2006 Q9 [11]}}