| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Intersection of two lines |
| Difficulty | Moderate -0.8 This is a straightforward C1 coordinate geometry question requiring standard techniques: finding gradient from two points, converting to general form, then solving simultaneous equations. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the arithmetic involved. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{12-3}{11-(-1)} = \frac{3}{4}\) | M1, A1 | M1: Correct method for the gradient. A1: Any correct fraction or decimal |
| \(y - 3 = \frac{3}{4}(x+1)\) or \(y - 12 = \frac{3}{4}(x-11)\) or \(y = \frac{3}{4}x + c\) with attempt at substitution to find \(c\) | M1 | Correct straight line method using either of the given points and a numerical gradient |
| \(4y - 3x - 15 = 0\) | A1 | Or equivalent with integer coefficients (\(= 0\) is required). This A1 should only be awarded in (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Solves their equation from part (a) and \(L_2\) simultaneously to eliminate one variable | M1 | Must reach as far as an equation in \(x\) only or in \(y\) only. (Allow slips in the algebra) |
| \(x = 3\) or \(y = 6\) | A1 | One of \(x=3\) or \(y=6\) |
| Both \(x = 3\) and \(y = 6\) | A1 | Values can be un-simplified fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((-1,3) \to -a + 3b + c = 0\) and \((11,12) \to 11a + 12b + c = 0\) | M1 | Substitutes the coordinates to obtain two equations |
| \(\therefore a = -\frac{3}{4}b,\ b = -\frac{4}{15}c\) | A1 | Obtains sufficient equations to establish values for \(a\), \(b\) and \(c\) |
| e.g. \(c = 1 \Rightarrow b = -\frac{4}{15},\ a = \frac{3}{15}\) | M1 | Obtains values for \(a\), \(b\) and \(c\) |
| \(\frac{3}{15}x - \frac{4}{15}y + 1 = 0 \Rightarrow 4y - 3x - 15 = 0\) | A1 | Correct equation |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{12-3}{11-(-1)} = \frac{3}{4}$ | M1, A1 | M1: Correct method for the gradient. A1: Any correct fraction or decimal |
| $y - 3 = \frac{3}{4}(x+1)$ or $y - 12 = \frac{3}{4}(x-11)$ or $y = \frac{3}{4}x + c$ with attempt at substitution to find $c$ | M1 | Correct straight line method using either of the given points and a numerical gradient |
| $4y - 3x - 15 = 0$ | A1 | Or equivalent with integer coefficients ($= 0$ **is** required). This A1 should only be awarded in (a) |
**Total: (4)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Solves their equation from part (a) and $L_2$ simultaneously to eliminate one variable | M1 | Must reach as far as an equation in $x$ only or in $y$ only. (Allow slips in the algebra) |
| $x = 3$ or $y = 6$ | A1 | One of $x=3$ or $y=6$ |
| **Both** $x = 3$ **and** $y = 6$ | A1 | Values can be un-simplified fractions |
*Fully correct answers with no working can score 3/3 in (b)*
**Total: (3) [7]**
**Way 2 (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(-1,3) \to -a + 3b + c = 0$ and $(11,12) \to 11a + 12b + c = 0$ | M1 | Substitutes the coordinates to obtain two equations |
| $\therefore a = -\frac{3}{4}b,\ b = -\frac{4}{15}c$ | A1 | Obtains sufficient equations to establish values for $a$, $b$ and $c$ |
| e.g. $c = 1 \Rightarrow b = -\frac{4}{15},\ a = \frac{3}{15}$ | M1 | Obtains values for $a$, $b$ and $c$ |
| $\frac{3}{15}x - \frac{4}{15}y + 1 = 0 \Rightarrow 4y - 3x - 15 = 0$ | A1 | Correct equation |
**Total: (4) [7]**6. The straight line $L _ { 1 }$ passes through the points $( - 1,3 )$ and $( 11,12 )$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for $L _ { 1 }$ in the form $a x + b y + c = 0$, where $a$, $b$ and $c$ are integers.
The line $L _ { 2 }$ has equation $3 y + 4 x - 30 = 0$.
\item Find the coordinates of the point of intersection of $L _ { 1 }$ and $L _ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2013 Q6 [7]}}