| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Moderate -0.8 This is a straightforward C1 coordinate geometry question requiring routine procedures: rearranging to find gradient, applying the perpendicular gradient rule (negative reciprocal), and using point-slope form. All steps are standard textbook exercises with no problem-solving or insight required, making it easier than average but not trivial since it involves multiple connected steps. |
| 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 | Mark | Guidance |
| \(4x + 2y - 3 = 0 \Rightarrow y = -2x + \frac{3}{2}\) | M1 | Attempt to write in the form \(y =\) |
| gradient \(= -2\) | A1 | Accept any unsimplified form, allow even with an incorrect value of "\(c\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4 + 2\frac{dy}{dx} = 0\) | M1 | Attempt to differentiate. Allow \(p \pm q\frac{dy}{dx} = 0,\ p,q \neq 0\) |
| gradient \(= -2\) | A1 | Accept any unsimplified form. Answer only scores M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using \(m_N = -\frac{1}{m_T}\) | M1 | Attempt to use \(m_N = -\frac{1}{\text{gradient from }(a)}\) |
| \(y - 5 = \frac{1}{2}(x-2)\) or uses \(y = mx+c\) to find \(c\) | M1 | Correct straight line method using a 'changed' gradient and the point \((2,5)\) |
| \(y = \frac{1}{2}x + 4\) | A1 | cao (isw) |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4x + 2y - 3 = 0 \Rightarrow y = -2x + \frac{3}{2}$ | M1 | Attempt to write in the form $y =$ |
| gradient $= -2$ | A1 | Accept any unsimplified form, allow even with an incorrect value of "$c$" |
**Way 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4 + 2\frac{dy}{dx} = 0$ | M1 | Attempt to differentiate. Allow $p \pm q\frac{dy}{dx} = 0,\ p,q \neq 0$ |
| gradient $= -2$ | A1 | Accept any unsimplified form. Answer only scores M1A1 |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $m_N = -\frac{1}{m_T}$ | M1 | Attempt to use $m_N = -\frac{1}{\text{gradient from }(a)}$ |
| $y - 5 = \frac{1}{2}(x-2)$ or uses $y = mx+c$ to find $c$ | M1 | Correct straight line method using a 'changed' gradient and the point $(2,5)$ |
| $y = \frac{1}{2}x + 4$ | A1 | cao (isw) |
---4. The line $L _ { 1 }$ has equation $4 x + 2 y - 3 = 0$
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of $L _ { 1 }$.
The line $L _ { 2 }$ is perpendicular to $L _ { 1 }$ and passes through the point $( 2,5 )$.
\item Find the equation of $L _ { 2 }$ in the form $y = m x + c$, where $m$ and $c$ are constants.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2013 Q4 [5]}}