| Exam Board | Edexcel |
|---|---|
| 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 using y=mx+c, finding intersections, verifying perpendicularity using gradients, and calculating triangle area. While it has multiple parts (14 marks total), each step follows textbook methods with no novel insight required. Slightly above average difficulty due to length and the perpendicularity proof, but well within typical C1 scope. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(y - 4 = 3(x + 6)\) | M1 | |
| \(y = 3x + 22\) | A1 | |
| (b) 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 | |
| (c) grad \(AB = \frac{2-4}{0-(-6)} = -\frac{1}{3}\) | M1 A1 | |
| grad \(AC = \frac{1-4}{-7-(-6)} = 3\) | ||
| grad \(AB \times\) grad \(AC = -\frac{1}{3} \times 3 = -1\) | M1 | |
| \(\therefore AB\) perp to \(AC \therefore \angle BAC = 90°\) | A1 | |
| (d) \(AB = \sqrt{(0+6)^2 + (2-4)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}\) | M1 A1 | |
| \(AC = \sqrt{(-7+6)^2 + (1-4)^2} = \sqrt{1 + 9} = \sqrt{10}\) | ||
| area \(= \frac{1}{2} \times 2\sqrt{10} \times \sqrt{10} = 10\) | M1 A1 | (14) |
**(a)** $y - 4 = 3(x + 6)$ | M1 |
$y = 3x + 22$ | A1 |
**(b)** 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 |
**(c)** grad $AB = \frac{2-4}{0-(-6)} = -\frac{1}{3}$ | M1 A1 |
grad $AC = \frac{1-4}{-7-(-6)} = 3$ | |
grad $AB \times$ grad $AC = -\frac{1}{3} \times 3 = -1$ | M1 |
$\therefore AB$ perp to $AC \therefore \angle BAC = 90°$ | A1 |
**(d)** $AB = \sqrt{(0+6)^2 + (2-4)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$ | M1 A1 |
$AC = \sqrt{(-7+6)^2 + (1-4)^2} = \sqrt{1 + 9} = \sqrt{10}$ | |
area $= \frac{1}{2} \times 2\sqrt{10} \times \sqrt{10} = 10$ | M1 A1 | (14)
**Total: (75)**The straight line $l$ has gradient 3 and passes through the point $A(-6, 4)$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for $l$ in the form $y = mx + c$. [2]
\end{enumerate}
The straight line $m$ has the equation $x - 7y + 14 = 0$.
Given that $m$ crosses the $y$-axis at the point $B$ and intersects $l$ at the point $C$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the coordinates of $B$ and $C$, [4]
\item show that $\angle BAC = 90°$, [4]
\item find the area of triangle $ABC$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q10 [14]}}