| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Parallel line through point |
| Difficulty | Easy -1.2 This is a straightforward application of parallel lines (same gradient) and point substitution. Students rearrange the given line to find m = -2/3, then use y - y₁ = m(x - x₁) with the given point. It's routine coordinate geometry with no problem-solving required, making it easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| \(2x + 3y = 6 \Rightarrow y = -\frac{2}{3}x + \ldots\) | B1 | States or implies gradient of \(l_1\) is \(-\frac{2}{3}\). Alternatively accept \(l_1\) written in form \(y = -\frac{2}{3}x + \ldots\) |
| Equation of \(l_2\) is \(y = -\frac{2}{3}x + c\) | M1 | States or implies gradient of \(l_2\) is the same as \(l_1\). If gradient of \(l_2\) is incorrect, must see evidence it has been linked with gradient of \(l_1\) |
| Substitutes \((3, -5)\) into \(y = -\frac{2}{3}x + c \Rightarrow -5 = -\frac{2}{3}\times 3 + c\) | M1 | Substitutes \((3,-5)\) into their \(y = -\frac{2}{3}x + c\). Also score for \(-\frac{2}{3} = \frac{y--5}{x-3}\) |
| \(y = -\frac{2}{3}x - 3\) | A1 | Accept forms like \(y = \left(-\frac{2}{3}\right)x + (-3)\) |
| Answer | Marks |
|---|---|
| Equation of \(l_2\) is \(2x + 3y = c\) | B1 M1 |
| Substitutes \((3,-5)\): \(6 - 15 = c\), giving \(2x + 3y = -9\) | M1 |
| \(y = -\frac{2}{3}x - 3\) | A1 |
## Question 3:
| $2x + 3y = 6 \Rightarrow y = -\frac{2}{3}x + \ldots$ | B1 | States or implies gradient of $l_1$ is $-\frac{2}{3}$. Alternatively accept $l_1$ written in form $y = -\frac{2}{3}x + \ldots$ |
| Equation of $l_2$ is $y = -\frac{2}{3}x + c$ | M1 | States or implies gradient of $l_2$ is the same as $l_1$. If gradient of $l_2$ is incorrect, must see evidence it has been linked with gradient of $l_1$ |
| Substitutes $(3, -5)$ into $y = -\frac{2}{3}x + c \Rightarrow -5 = -\frac{2}{3}\times 3 + c$ | M1 | Substitutes $(3,-5)$ into their $y = -\frac{2}{3}x + c$. Also score for $-\frac{2}{3} = \frac{y--5}{x-3}$ |
| $y = -\frac{2}{3}x - 3$ | A1 | Accept forms like $y = \left(-\frac{2}{3}\right)x + (-3)$ |
**Alternative:**
| Equation of $l_2$ is $2x + 3y = c$ | B1 M1 | |
| Substitutes $(3,-5)$: $6 - 15 = c$, giving $2x + 3y = -9$ | M1 | |
| $y = -\frac{2}{3}x - 3$ | A1 | |
---\begin{enumerate}
\item The line $l _ { 1 }$ has equation $2 x + 3 y = 6$
\end{enumerate}
The line $l _ { 2 }$ is parallel to the line $l _ { 1 }$ and passes through the point $( 3 , - 5 )$.\\
Find the equation for the line $l _ { 2 }$ in the form $y = m x + c$, where $m$ and $c$ are constants.
\hfill \mbox{\textit{Edexcel C12 2017 Q3 [4]}}