| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Equation of line through two points |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing two standard procedures: finding the equation of a line through two points and calculating distance between points. Both parts require direct application of formulas with minimal problem-solving, 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=01.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(m = \frac{4-(-3)}{-6-8}\) or \(\frac{-3-4}{8-(-6)}\), \(= \frac{-3-4}{8-(-6)}\), \(= \frac{-7}{14}\) or \(\frac{7}{-14}\) (= \(-\frac{1}{2}\)) | M1, A1 | |
| Equation: \(y - 4 = -\frac{1}{2}(x - (-6))\) or \(y - (-3) = -\frac{1}{2}(x - 8)\) | M1 | |
| \(x + 2y - 2 = 0\) (or equiv. with integer coefficients... must have '= 0') | A1 | (e.g. 14\(y\) + 7\(x\) - 14 = 0 and 14 - 7\(x\) - 14\(y\) = 0 are acceptable) |
| Answer | Marks |
|---|---|
| (b) \((-6-8)^2 + (4-(-3))^2\) | M1 |
| \(14^2 + 7^2\) or \((-14)^2 + 7^2\) or \(14^2 + (-7)^2\) (M1 A1 may be implied by 245) | A1 |
| \(AB = \sqrt{14^2 + 7^2}\) or \(\sqrt{7^2(2^2+1^2)}\) or \(\sqrt{245}\) | A1 |
| \(7\sqrt{5}\) | A1also |
(a) $m = \frac{4-(-3)}{-6-8}$ or $\frac{-3-4}{8-(-6)}$, $= \frac{-3-4}{8-(-6)}$, $= \frac{-7}{14}$ or $\frac{7}{-14}$ (= $-\frac{1}{2}$) | M1, A1 |
Equation: $y - 4 = -\frac{1}{2}(x - (-6))$ or $y - (-3) = -\frac{1}{2}(x - 8)$ | M1 |
$x + 2y - 2 = 0$ (or equiv. with integer coefficients... must have '= 0') | A1 | (e.g. 14$y$ + 7$x$ - 14 = 0 and 14 - 7$x$ - 14$y$ = 0 are acceptable)
**Total for (a): 4 marks**
(b) $(-6-8)^2 + (4-(-3))^2$ | M1 |
$14^2 + 7^2$ or $(-14)^2 + 7^2$ or $14^2 + (-7)^2$ (M1 A1 may be implied by 245) | A1 |
$AB = \sqrt{14^2 + 7^2}$ or $\sqrt{7^2(2^2+1^2)}$ or $\sqrt{245}$ | A1 |
$7\sqrt{5}$ | A1also |
**Total for (b): 3 marks**
**Total: 7 marks**
---4. The point $A ( - 6,4 )$ and the point $B ( 8 , - 3 )$ lie on the line $L$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for $L$ in the form $a x + b y + c = 0$, where $a$, $b$ and $c$ are integers.
\item Find the distance $A B$, giving your answer in the form $k \sqrt { 5 }$, where $k$ is an integer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2008 Q4 [7]}}