| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | January |
| 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 basic skills: substituting a point into an equation to verify it lies on the line, finding the perpendicular gradient (negative reciprocal), and using point-slope form. These are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial since it requires multiple standard steps. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks |
|---|---|
| (a) \(y = 5 - (2 \times 3) = -1\) (or equivalent verification) | B1 |
| (b) Gradient of \(L\) is \(\frac{1}{2}\) | B1 |
| \(y - (-1) = \frac{1}{2}(x-3)\) (ft from a changed gradient) | M1 A1ft |
| \(x - 2y - 5 = 0\) (or equiv. with integer coefficients) | A1 |
| Answer | Marks |
|---|---|
| Guidance: (a) \(y - (-1) = -2(x-3) \Rightarrow y = 5 - 2x\) is fine for B1. Just a table of values including \(x = 3, y = -1\) is insufficient. (b) M1: eqn of a line through \((3, -1)\), with any numerical gradient (except 0 or \(\infty\)). For the M1 A1ft, the equation may be in any form, e.g. \(\frac{y-(-1)}{x-3} = \frac{1}{2}\). Alternatively, the M1 may be scored by using \(y = mx + c\) with a numerical gradient and substituting \((3, -1)\) to find the value of \(c\), with A1ft if the value of \(c\) follows through correctly from a changed gradient. Allow \(x - 2y = 5\) or equiv., but must be integer coefficients. The "=" 0" can be implied if correct working precedes. |
**(a)** $y = 5 - (2 \times 3) = -1$ (or equivalent verification) | B1 |
**(b)** Gradient of $L$ is $\frac{1}{2}$ | B1 |
$y - (-1) = \frac{1}{2}(x-3)$ (ft from a changed gradient) | M1 A1ft |
$x - 2y - 5 = 0$ (or equiv. with integer coefficients) | A1 |
**Total: 5 marks**
| Guidance: (a) $y - (-1) = -2(x-3) \Rightarrow y = 5 - 2x$ is fine for B1. Just a table of values including $x = 3, y = -1$ is insufficient. (b) M1: eqn of a line through $(3, -1)$, with any numerical gradient (except 0 or $\infty$). For the M1 A1ft, the equation may be in any form, e.g. $\frac{y-(-1)}{x-3} = \frac{1}{2}$. Alternatively, the M1 may be scored by using $y = mx + c$ with a numerical gradient and substituting $(3, -1)$ to find the value of $c$, with A1ft if the value of $c$ follows through correctly from a changed gradient. Allow $x - 2y = 5$ or equiv., but must be integer coefficients. The "=" 0" can be implied if correct working precedes.
---3. The line $L$ has equation $y = 5 - 2 x$.
\begin{enumerate}[label=(\alph*)]
\item Show that the point $P ( 3 , - 1 )$ lies on $L$.
\item Find an equation of the line perpendicular to $L$, which passes through $P$. Give your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2006 Q3 [5]}}