| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| 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 C1 techniques: finding gradient from equation, parallel lines, intersection of lines, and distance formula. Part (c) requires solving a quadratic from the distance condition, which adds slight challenge, but all parts are routine textbook exercises with no novel problem-solving required. Slightly easier than average due to straightforward application of formulas. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{1}{2}(7 - 3x)\) | M1 | attempt at \(y = \ldots\) or use of 2 correct points using \(\frac{\Delta y}{\Delta x}\) |
| \(\Rightarrow\) gradient \(= -\frac{3}{2}\) | A1 | condone slip in rearranging if gradient is correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(y =\) their grad \(\cdot x + c\) and substitution of \(x = 2, y = -7\) | M1 | or using \(3x + 2y = k\) with \(x = 2, y = -7\) and attempt to find \(k\); or \(y - (-7) =\) their grad \(\cdot (x - 2)\) |
| \(y = -\frac{3}{2}x + c, c = -4\) | A1 | correct equation in any form: \(y + 7 = -\frac{3}{2}(x - 2), 3x + 2y + 8 = 0\), etc |
| \((x = 0 \Rightarrow) y = -4\) | A1cso | or y-intercept = -4 or D(0, -4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(3x + 2(1 - 4x) = 7, y = 1 - \frac{4}{3}(7 - 2y)\) | M1 | elimination of \(y\) (or x) (condone one slip) |
| \(x = -1\) | A1 | one coordinate correct |
| \(y = 5\) | A1 | other coordinate correct; coordinates of A(-1, 5) |
| Answer | Marks | Guidance |
|---|---|---|
| \((5 - 2)^2 + (k + 7)^2 = 5^2\) | M1 | condone one sign slip within one bracket |
| \((k + 7 = 4\) or \(k + 7 = -4)\) | A1 | one correct value of k |
| \(k = -3\) or \(k = -11\) | A1 | both correct (and no other values) |
**3(a)(i)**
$y = \frac{1}{2}(7 - 3x)$ | M1 | attempt at $y = \ldots$ or use of 2 correct points using $\frac{\Delta y}{\Delta x}$
$\Rightarrow$ gradient $= -\frac{3}{2}$ | A1 | condone slip in rearranging if gradient is correct | 2
**3(a)(ii)**
$y =$ their grad $\cdot x + c$ and substitution of $x = 2, y = -7$ | M1 | or using $3x + 2y = k$ with $x = 2, y = -7$ and attempt to find $k$; or $y - (-7) =$ their grad $\cdot (x - 2)$
$y = -\frac{3}{2}x + c, c = -4$ | A1 | correct equation in any form: $y + 7 = -\frac{3}{2}(x - 2), 3x + 2y + 8 = 0$, etc
$(x = 0 \Rightarrow) y = -4$ | A1cso | or y-intercept = -4 or D(0, -4) | 3
**3(b)**
$3x + 2(1 - 4x) = 7, y = 1 - \frac{4}{3}(7 - 2y)$ | M1 | elimination of $y$ (or x) (condone one slip)
$x = -1$ | A1 | one coordinate correct
$y = 5$ | A1 | other coordinate correct; coordinates of A(-1, 5) | 3
**3(c)**
$(5 - 2)^2 + (k + 7)^2 = 5^2$ | M1 | condone one sign slip within one bracket
$(k + 7 = 4$ or $k + 7 = -4)$ | A1 | one correct value of k
$k = -3$ or $k = -11$ | A1 | both correct (and no other values) | 3
---3 The line $A B$ has equation $3 x + 2 y = 7$. The point $C$ has coordinates $( 2 , - 7 )$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the gradient of $A B$.
\item The line which passes through $C$ and which is parallel to $A B$ crosses the $y$-axis at the point $D$. Find the $y$-coordinate of $D$.
\end{enumerate}\item The line with equation $y = 1 - 4 x$ intersects the line $A B$ at the point $A$. Find the coordinates of $A$.
\item The point $E$ has coordinates $( 5 , k )$. Given that $C E$ has length 5 , find the two possible values of the constant $k$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2011 Q3 [11]}}