| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| 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 two-part question testing basic coordinate geometry: rearranging a linear equation to find gradient, then using y - y₁ = m(x - x₁) with a parallel line. Both parts are routine textbook exercises requiring only direct application of standard formulas with no problem-solving or insight needed. |
| 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 | Mark | Guidance |
| Gradient is \(5\) | B1 | Accept \(5\) or \(\frac{10}{2}\) or exact equivalent. Do not accept if embedded within an equation. Must see \(5\) (or equivalent). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Gradient of parallel line equals gradient from (a) | M1 | Implied by sight of their '\(5\)' in a gradient equation. e.g. \(y = \text{'5'}x + c\) |
| \(y - \frac{4}{3} = \text{"5"}\!\left(x - \left(-\frac{1}{3}\right)\right)\) | M1 | Attempt to find equation using \(\left(-\frac{1}{3}, \frac{4}{3}\right)\) and a numerical gradient. Accept \(y - \frac{4}{3} = \text{"numerical }m\text{"}\!\left(x - -\frac{1}{3}\right)\). Allow one sign slip for the coordinate. If \(y = mx + c\) used, must proceed to finding \(c\). |
| \(y = 5x + 3\) | A1 | Correct answer only (no equivalents). Allow \(y = 5x + c\) followed by \(c = 3\). |
## Question 1:
**(a)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Gradient is $5$ | B1 | Accept $5$ or $\frac{10}{2}$ or exact equivalent. Do not accept if embedded within an equation. Must see $5$ (or equivalent). |
**(b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Gradient of parallel line equals gradient from (a) | M1 | Implied by sight of their '$5$' in a gradient equation. e.g. $y = \text{'5'}x + c$ |
| $y - \frac{4}{3} = \text{"5"}\!\left(x - \left(-\frac{1}{3}\right)\right)$ | M1 | Attempt to find equation using $\left(-\frac{1}{3}, \frac{4}{3}\right)$ and a numerical gradient. Accept $y - \frac{4}{3} = \text{"numerical }m\text{"}\!\left(x - -\frac{1}{3}\right)$. Allow one sign slip for the coordinate. If $y = mx + c$ used, must proceed to finding $c$. |
| $y = 5x + 3$ | A1 | Correct answer only (no equivalents). Allow $y = 5x + c$ followed by $c = 3$. |
---\begin{enumerate}
\item The line $l _ { 1 }$ has equation
\end{enumerate}
$$10 x - 2 y + 7 = 0$$
(a) Find the gradient of $l _ { 1 }$
The line $l _ { 2 }$ is parallel to the line $l _ { 1 }$ and passes through the point $\left( - \frac { 1 } { 3 } , \frac { 4 } { 3 } \right)$.\\
(b) Find the equation of $l _ { 2 }$ in the form $y = m x + c$, where $m$ and $c$ are constants.\\
\hfill \mbox{\textit{Edexcel C12 2015 Q1 [4]}}