| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 11 |
| 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 standard C1 coordinate geometry question requiring gradient calculation, perpendicular line equations, and distance verification. While it has three parts and requires multiple techniques (gradient, equation forms, distance formula), each step follows routine procedures with no novel insight needed. The 'show' in part (iii) makes it slightly more demanding than pure recall, but the path is clear and algorithmic. |
| 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 | Notes |
| \(\text{grad} = \frac{1-5}{4-(-2)} = -\frac{2}{3}\) | M1 A1 | |
| \(\therefore y - 5 = -\frac{2}{3}(x+2)\) | M1 | |
| \(3y - 15 = -2x - 4\) | ||
| \(2x + 3y = 11\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\text{grad } l_2 = \frac{-1}{-\frac{2}{3}} = \frac{3}{2}\) | M1 A1 | |
| \(\therefore y - 1 = \frac{3}{2}(x-4)\) \([3x - 2y = 10]\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| At \(C\), \(x = 0\) \(\therefore y = -5\) \(\Rightarrow\) \(C(0,-5)\) | B1 | |
| \(AB = \sqrt{(4+2)^2 + (1-5)^2} = \sqrt{36+16} = \sqrt{52}\) | M1 A1 | |
| \(BC = \sqrt{(0-4)^2 + (-5-1)^2} = \sqrt{16+36} = \sqrt{52}\) | ||
| \(AB = BC\) \(\therefore\) triangle \(ABC\) is isosceles | A1 | (11) |
# Question 9:
## Part (i)
| Answer/Working | Marks | Notes |
|---|---|---|
| $\text{grad} = \frac{1-5}{4-(-2)} = -\frac{2}{3}$ | M1 A1 | |
| $\therefore y - 5 = -\frac{2}{3}(x+2)$ | M1 | |
| $3y - 15 = -2x - 4$ | | |
| $2x + 3y = 11$ | A1 | |
## Part (ii)
| Answer/Working | Marks | Notes |
|---|---|---|
| $\text{grad } l_2 = \frac{-1}{-\frac{2}{3}} = \frac{3}{2}$ | M1 A1 | |
| $\therefore y - 1 = \frac{3}{2}(x-4)$ $[3x - 2y = 10]$ | A1 | |
## Part (iii)
| Answer/Working | Marks | Notes |
|---|---|---|
| At $C$, $x = 0$ $\therefore y = -5$ $\Rightarrow$ $C(0,-5)$ | B1 | |
| $AB = \sqrt{(4+2)^2 + (1-5)^2} = \sqrt{36+16} = \sqrt{52}$ | M1 A1 | |
| $BC = \sqrt{(0-4)^2 + (-5-1)^2} = \sqrt{16+36} = \sqrt{52}$ | | |
| $AB = BC$ $\therefore$ triangle $ABC$ is isosceles | A1 | **(11)** |
---9. The straight line $l _ { 1 }$ passes through the point $A ( - 2,5 )$ and the point $B ( 4,1 )$.\\
(i) Find an equation for $l _ { 1 }$ in the form $a x + b y = c$, where $a , b$ and $c$ are integers.
The straight line $l _ { 2 }$ passes through $B$ and is perpendicular to $l _ { 1 }$.\\
(ii) Find an equation for $l _ { 2 }$.
Given that $l _ { 2 }$ meets the $y$-axis at the point $C$,\\
(iii) show that triangle $A B C$ is isosceles.\\
\hfill \mbox{\textit{OCR C1 Q9 [11]}}