| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Parameter from distance condition |
| Difficulty | Standard +0.3 This is a standard C1 coordinate geometry question covering routine techniques: gradient, line equations, midpoint, perpendicular lines, and distance formula with parameter. Part (c) requires solving a quadratic from the distance condition, which is slightly beyond pure recall but still a textbook exercise with no novel insight needed. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.10e Position vectors: and displacement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(m = \frac{-5-2}{3-(-1)} = \frac{-7}{4}\) | M1 | Correct use of gradient formula |
| \(m = -\frac{7}{4}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y - 2 = -\frac{7}{4}(x+1)\) | M1 | Correct method using point and gradient |
| \(4y - 8 = -7x - 7\) | M1 | Attempt to rearrange to required form |
| \(7x + 4y = 1\) | A1 | Must be integer form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(M = (1, -\frac{3}{2})\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Perpendicular gradient \(= \frac{4}{7}\) | B1 | |
| \(y + \frac{3}{2} = \frac{4}{7}(x-1)\) | M1 | Using \(M\) and perpendicular gradient |
| \(7y + \frac{21}{2} = 4x - 4\) or equivalent | A1 | Any correct form e.g. \(4x - 7y = \frac{29}{2}\) or \(8x - 14y = 29\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((k-(-1))^2 + (2k+3-2)^2 = 13\) | M1 | Correct distance formula squared |
| \((k+1)^2 + (2k+1)^2 = 13\) | A1 | |
| \(k^2 + 2k + 1 + 4k^2 + 4k + 1 = 13\) | M1 | Expand and simplify |
| \(5k^2 + 6k - 11 = 0\) | ||
| \((5k + 11)(k-1) = 0\) | M1 | Solve quadratic |
| \(k = 1\) or \(k = -\frac{11}{5}\) | A1 | Both values required |
# Question 1:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $m = \frac{-5-2}{3-(-1)} = \frac{-7}{4}$ | M1 | Correct use of gradient formula |
| $m = -\frac{7}{4}$ | A1 | |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $y - 2 = -\frac{7}{4}(x+1)$ | M1 | Correct method using point and gradient |
| $4y - 8 = -7x - 7$ | M1 | Attempt to rearrange to required form |
| $7x + 4y = 1$ | A1 | Must be integer form |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $M = (1, -\frac{3}{2})$ | B1 | |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Perpendicular gradient $= \frac{4}{7}$ | B1 | |
| $y + \frac{3}{2} = \frac{4}{7}(x-1)$ | M1 | Using $M$ and perpendicular gradient |
| $7y + \frac{21}{2} = 4x - 4$ or equivalent | A1 | Any correct form e.g. $4x - 7y = \frac{29}{2}$ or $8x - 14y = 29$ |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $(k-(-1))^2 + (2k+3-2)^2 = 13$ | M1 | Correct distance formula squared |
| $(k+1)^2 + (2k+1)^2 = 13$ | A1 | |
| $k^2 + 2k + 1 + 4k^2 + 4k + 1 = 13$ | M1 | Expand and simplify |
| $5k^2 + 6k - 11 = 0$ | | |
| $(5k + 11)(k-1) = 0$ | M1 | Solve quadratic |
| $k = 1$ or $k = -\frac{11}{5}$ | A1 | Both values required |
---1 The point $A$ has coordinates $( - 1,2 )$ and the point $B$ has coordinates $( 3 , - 5 )$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the gradient of $A B$.
\item Hence find an equation of the line $A B$, giving your answer in the form $p x + q y = r$, where $p , q$ and $r$ are integers.
\end{enumerate}\item The midpoint of $A B$ is $M$.
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of $M$.
\item Find an equation of the line which passes through $M$ and which is perpendicular to $A B$. [3 marks]
\end{enumerate}\item The point $C$ has coordinates $( k , 2 k + 3 )$. Given that the distance from $A$ to $C$ is $\sqrt { 13 }$, find the two possible values of the constant $k$.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2014 Q1 [13]}}