| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Equation of line through two points |
| Difficulty | Easy -1.2 This is a straightforward C1 coordinate geometry question requiring only direct application of standard formulas: midpoint formula and gradient calculation followed by y = mx + c form. Both parts are routine textbook exercises with no problem-solving or conceptual challenge beyond basic recall and arithmetic. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 |
| Answer | Marks | Guidance |
|---|---|---|
| Midpoint \(= \left(\frac{1+5}{2}, \frac{2+8}{2}\right) = (3, 5)\) | B1 B1 | B1 each coordinate |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient \(m = \frac{8-2}{5-1} = \frac{6}{4} = \frac{3}{2}\) | M1 A1 | M1 for correct gradient method |
| \(y - 2 = \frac{3}{2}(x-1)\) | M1 | M1 for line equation using their gradient and a given point |
| \(y = \frac{3}{2}x + \frac{1}{2}\) | A1 |
# Question 3:
## Part (a):
| Midpoint $= \left(\frac{1+5}{2}, \frac{2+8}{2}\right) = (3, 5)$ | B1 B1 | B1 each coordinate |
## Part (b):
| Gradient $m = \frac{8-2}{5-1} = \frac{6}{4} = \frac{3}{2}$ | M1 A1 | M1 for correct gradient method |
| $y - 2 = \frac{3}{2}(x-1)$ | M1 | M1 for line equation using their gradient and a given point |
| $y = \frac{3}{2}x + \frac{1}{2}$ | A1 | |
---3. The points $A$ and $B$ have coordinates $( 1,2 )$ and $( 5,8 )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the mid-point of $A B$.
\item Find, in the form $y = m x + c$, an equation for the straight line through $A$ and $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q3 [6]}}