| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| 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 routine application of gradient formulas and perpendicular line properties. Part (a) involves simple rearrangement to find gradient, part (b) uses the standard perpendicular gradient rule (negative reciprocal) and point-slope form. Both are textbook exercises with no problem-solving or insight 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Put equation in form \(y=mx(+c)\) and attempt to extract \(m\) or \(mx\); or find 2 points and use gradient formula | M1 | Condone sign errors and ignore \(c\) for M mark |
| Gradient \(= -\frac{3}{5}\) (or equivalent) | A1 | Answer only: \(-\frac{3}{5}\) scores M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient of perpendicular line \(= \frac{-1}{``(-3/5)"}\) using \(-\frac{1}{m}\) with \(m\) from part (a) | M1 | |
| \(y-1 = ``\left(\frac{5}{3}\right)"(x-3)\) | M1 | 2nd M: equation of straight line through \((3,1)\) with any numerical gradient (except 0 or \(\infty\)) |
| \(y = \frac{5}{3}x - 4\) (must be in this form; allow \(y=\frac{5}{3}x-\frac{12}{3}\) but not \(y=\frac{5x-12}{3}\)) | A1 | This A mark is dependent upon both M marks |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Put equation in form $y=mx(+c)$ and attempt to extract $m$ or $mx$; or find 2 points and use gradient formula | M1 | Condone sign errors and ignore $c$ for M mark |
| Gradient $= -\frac{3}{5}$ (or equivalent) | A1 | Answer only: $-\frac{3}{5}$ scores M1 A1 |
**Subtotal: (2)**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of perpendicular line $= \frac{-1}{``(-3/5)"}$ using $-\frac{1}{m}$ with $m$ from part (a) | M1 | |
| $y-1 = ``\left(\frac{5}{3}\right)"(x-3)$ | M1 | 2nd M: equation of straight line through $(3,1)$ with any numerical gradient (except 0 or $\infty$) |
| $y = \frac{5}{3}x - 4$ (must be in this form; allow $y=\frac{5}{3}x-\frac{12}{3}$ but not $y=\frac{5x-12}{3}$) | A1 | This A mark is dependent upon both M marks |
**Subtotal: (3)**
**Total: [5]**
---The line $l _ { 1 }$ has equation $3 x + 5 y - 2 = 0$
\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 $( 3,1 )$.
\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 2010 Q3 [5]}}