| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Intersection of two lines |
| Difficulty | Standard +0.3 This is a standard multi-part coordinate geometry question requiring routine techniques: finding line equations, intersections, verifying perpendicularity via gradients, and calculating triangle area. While it has four parts worth several marks, each step follows textbook methods with no novel insight required, 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 |
|---|---|
| \(y - 4 = 3(x+6)\) | M1 |
| \(y = 3x + 22\) | A1 |
| Answer | Marks |
|---|---|
| At \(B\), \(x=0 \therefore y=2 \Rightarrow B(0,2)\) | B1 |
| At \(C\), \(x - 7(3x+22) + 14 = 0\) | M1 |
| \(x = -7\) | A1 |
| \(\therefore C(-7,1)\) | A1 |
| Answer | Marks |
|---|---|
| \(\text{grad } AB = \frac{2-4}{0-(-6)} = -\frac{1}{3}\) | M1 A1 |
| \(\text{grad } AC = \frac{1-4}{-7-(-6)} = 3\) | |
| \(\text{grad } AB \times \text{grad } AC = -\frac{1}{3} \times 3 = -1\) | M1 |
| \(\therefore AB \perp AC \therefore \angle BAC = 90°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB = \sqrt{(0+6)^2+(2-4)^2} = \sqrt{36+4} = \sqrt{40} = 2\sqrt{10}\) | M1 A1 | |
| \(AC = \sqrt{(-7+6)^2+(1-4)^2} = \sqrt{1+9} = \sqrt{10}\) | ||
| \(\text{area} = \frac{1}{2} \times 2\sqrt{10} \times \sqrt{10} = 10\) | M1 A1 | (14) |
# Question 10:
## Part (i):
| $y - 4 = 3(x+6)$ | M1 | |
|---|---|---|
| $y = 3x + 22$ | A1 | |
## Part (ii):
| At $B$, $x=0 \therefore y=2 \Rightarrow B(0,2)$ | B1 | |
|---|---|---|
| At $C$, $x - 7(3x+22) + 14 = 0$ | M1 | |
| $x = -7$ | A1 | |
| $\therefore C(-7,1)$ | A1 | |
## Part (iii):
| $\text{grad } AB = \frac{2-4}{0-(-6)} = -\frac{1}{3}$ | M1 A1 | |
|---|---|---|
| $\text{grad } AC = \frac{1-4}{-7-(-6)} = 3$ | | |
| $\text{grad } AB \times \text{grad } AC = -\frac{1}{3} \times 3 = -1$ | M1 | |
| $\therefore AB \perp AC \therefore \angle BAC = 90°$ | A1 | |
## Part (iv):
| $AB = \sqrt{(0+6)^2+(2-4)^2} = \sqrt{36+4} = \sqrt{40} = 2\sqrt{10}$ | M1 A1 | |
|---|---|---|
| $AC = \sqrt{(-7+6)^2+(1-4)^2} = \sqrt{1+9} = \sqrt{10}$ | | |
| $\text{area} = \frac{1}{2} \times 2\sqrt{10} \times \sqrt{10} = 10$ | M1 A1 | **(14)** |
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**Total: (72)**10. The straight line $l$ has gradient 3 and passes through the point $A ( - 6,4 )$.\\
(i) Find an equation for $l$ in the form $y = m x + c$.
The straight line $m$ has the equation $x - 7 y + 14 = 0$.\\
Given that $m$ crosses the $y$-axis at the point $B$ and intersects $l$ at the point $C$,\\
(ii) find the coordinates of $B$ and $C$,\\
(iii) show that $\angle B A C = 90 ^ { \circ }$,\\
(iv) find the area of triangle $A B C$.
\hfill \mbox{\textit{OCR C1 Q10 [14]}}