| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Coordinates from geometric constraints |
| Difficulty | Moderate -0.8 This is a straightforward C1 coordinate geometry question requiring standard techniques: finding gradient from two points, converting to general form, and using the perpendicular gradient condition. Both parts are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation required. |
| 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) \(\text{grad} = \frac{-4-6}{1-(-3)} = -\frac{5}{3}\) | M1 A1 | |
| \(\therefore y - 6 = -\frac{5}{3}(x+3)\) | M1 | |
| \(2y - 12 = -5x - 15\) | ||
| \(5x + 2y + 3 = 0\) | A1 | |
| (b) \(m: y = -\frac{2}{k}x - \frac{7}{k}\) \(\therefore \text{grad} = -\frac{2}{k}\) | M1 A1 | |
| \(l\) and \(m\) perp. \(\therefore -\frac{5}{3} \times -\frac{2}{k} = -1\) | M1 | |
| \(k = -5\) | A1 | (8) |
**(a)** $\text{grad} = \frac{-4-6}{1-(-3)} = -\frac{5}{3}$ | M1 A1 |
$\therefore y - 6 = -\frac{5}{3}(x+3)$ | M1 |
$2y - 12 = -5x - 15$ | |
$5x + 2y + 3 = 0$ | A1 |
**(b)** $m: y = -\frac{2}{k}x - \frac{7}{k}$ $\therefore \text{grad} = -\frac{2}{k}$ | M1 A1 |
$l$ and $m$ perp. $\therefore -\frac{5}{3} \times -\frac{2}{k} = -1$ | M1 |
$k = -5$ | A1 | (8)The straight line $l$ passes through the point $P(-3, 6)$ and the point $Q(1, -4)$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for $l$ in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are integers. [4]
\end{enumerate}
The straight line $m$ has the equation $2x + ky + 7 = 0$, where $k$ is a constant.
Given that $l$ and $m$ are perpendicular,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the value of $k$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q6 [8]}}