| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Parameter from distance condition |
| Difficulty | Moderate -0.3 This is a multi-part coordinate geometry question covering standard techniques: finding gradient from line equation, perpendicular lines, simultaneous equations for intersection, and distance formula with a parameter. All parts are routine C1 material requiring straightforward application of formulas rather than problem-solving insight. Slightly easier than average due to the step-by-step scaffolding, though part (c) requires solving a quadratic equation. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = -\frac{3}{5}x + \ldots ;\) Gradient \(AB = -\frac{3}{5}\) | M1 | Attempt to find \(y =\) or \(\Delta y/\Delta x\); or \(\frac{3}{5}\) or \(3x/5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of perpendicular \(= \frac{5}{3}\) | A1, M1, A1✓ | Gradient correct – condone slip in \(y = \ldots\); Stated or used correctly; ft gradient of \(AB\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow y + 2 = \frac{5}{3}(x - 6)\) | A1 | CSO Any correct form e.g. \(y = \frac{5}{3}x - 12\), \(5x - 3y = 36\) etc |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 7\) | M1, A1, A1 | Must use \(3x + 5y = 8; 2x + 3y = 3\); \(B(-9,7)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(k = 1\) or \(k = -5\) | M1, A1, A1 | Diagram with 3.4, 5 triangle; Condone slip in one term (or \(k + 2 = 3\)); SC1 with no working for spotting one correct value of \(k\). Full marks if both values spotted with no contradictory work |
## 2(a)(i)
$y = -\frac{3}{5}x + \ldots ;$ Gradient $AB = -\frac{3}{5}$ | M1 | Attempt to find $y =$ or $\Delta y/\Delta x$; or $\frac{3}{5}$ or $3x/5$
## 2(a)(ii)
$m_1m_2 = -1$
Gradient of perpendicular $= \frac{5}{3}$ | A1, M1, A1✓ | Gradient correct – condone slip in $y = \ldots$; Stated or used correctly; ft gradient of $AB$
## 2(b)
$\Rightarrow y + 2 = \frac{5}{3}(x - 6)$ | A1 | CSO Any correct form e.g. $y = \frac{5}{3}x - 12$, $5x - 3y = 36$ etc
Eliminating $x$ or $y$ (unsimplified)
$x = -9$
$y = 7$ | M1, A1, A1 | Must use $3x + 5y = 8; 2x + 3y = 3$; $B(-9,7)$
## 2(c)
$4^2 + (k + 2)^2 (= 25)$ or $16 + d^2 = 25$
$k = 1$ or $k = -5$ | M1, A1, A1 | Diagram with 3.4, 5 triangle; Condone slip in one term (or $k + 2 = 3$); SC1 with no working for spotting one correct value of $k$. Full marks if both values spotted with no contradictory work
**Total: 11**
---2 The line $A B$ has equation $3 x + 5 y = 8$ and the point $A$ has coordinates (6, -2).
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the gradient of $A B$.
\item Hence find an equation of the straight line which is perpendicular to $A B$ and which passes through $A$.
\end{enumerate}\item The line $A B$ intersects the line with equation $2 x + 3 y = 3$ at the point $B$. Find the coordinates of $B$.
\item The point $C$ has coordinates $( 2 , k )$ and the distance from $A$ to $C$ is 5 . Find the two possible values of the constant $k$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2007 Q2 [11]}}