| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| 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 two-part question testing basic coordinate geometry: finding a gradient from two points, then using the perpendicular gradient property (negative reciprocal) to write an equation. Both are standard textbook exercises requiring only routine application of formulas with no problem-solving 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 | Marks | Guidance |
| Gradient \(= \frac{4-(-8)}{-1-5} = -2\) | M1 A1 [2] | M1: attempts \(\frac{\Delta y}{\Delta x}\); condone one sign slip; A1: cao, do not allow fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Perpendicular gradient \(= \frac{-1}{m}\left(=\frac{1}{2}\right)\) | M1 | Uses or states negative reciprocal of their gradient |
| \(y-(-8) = \frac{1}{2}(x-5)\) or \(y = \frac{1}{2}x + c\), with \(c = -8 - \frac{1}{2}\times 5\) so \(y = \frac{1}{2}x - 10\frac{1}{2}\) | M1 A1 | M1: uses line equation with point \((5,-8)\) and changed gradient; condone one sign slip |
| \(x - 2y - 21 = 0\) | A1 [4] | cao; accept \(k(x-2y-21)=0\) where \(k\) is integer \(\neq 0\); allow \(1x - 2y - 21 = 0\) |
# Question 3:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient $= \frac{4-(-8)}{-1-5} = -2$ | M1 A1 [2] | M1: attempts $\frac{\Delta y}{\Delta x}$; condone one sign slip; A1: cao, do not allow fractions |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Perpendicular gradient $= \frac{-1}{m}\left(=\frac{1}{2}\right)$ | M1 | Uses or states negative reciprocal of their gradient |
| $y-(-8) = \frac{1}{2}(x-5)$ or $y = \frac{1}{2}x + c$, with $c = -8 - \frac{1}{2}\times 5$ so $y = \frac{1}{2}x - 10\frac{1}{2}$ | M1 A1 | M1: uses line equation with point $(5,-8)$ and changed gradient; condone one sign slip |
| $x - 2y - 21 = 0$ | A1 [4] | cao; accept $k(x-2y-21)=0$ where $k$ is integer $\neq 0$; allow $1x - 2y - 21 = 0$ |
---3. The line $l _ { 1 }$ passes through the points $A ( - 1,4 )$ and $B ( 5 , - 8 )$
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of $l _ { 1 }$
The line $l _ { 2 }$ is perpendicular to the line $l _ { 1 }$ and passes through the point $B ( 5 , - 8 )$
\item Find an equation for $l _ { 2 }$ in the form $a x + b y + c = 0$, where $a$, b and $c$ are integers.\\
II\\
"
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q3 [6]}}