| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Intersection of two lines |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard C1 coordinate geometry techniques: substituting a point into a line equation, finding gradient from standard form, using perpendicular gradients, and solving simultaneous linear equations. All parts are routine applications of basic formulas with no problem-solving insight required, making it slightly easier than average but not trivial due to the multiple parts and algebraic manipulation needed. |
| 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 |
|---|---|---|
| (a) \(3p - 4(p + 2) + 5 = 0\) | M1 | condone omission of brackets or one sign error |
| \(\Rightarrow p = -3\) | A1 | 2 marks total |
| (b) \(y = \frac{3}{4}x + \frac{5}{4}\) | M1 | rearranging into form \(y = \pm\frac{3}{4}x + c\) |
| (gradient \(AB =\) ) \(\frac{3}{4}\) | A1 | condone slips in rearranging if gradient is correct; 2 marks total |
| (c) (gradient \(AC =\) ) \(\frac{k-2}{-5-1}\) or \(\frac{2-k}{1-(-5)}\) | M1 | (condone one sign error) |
| "their" \(\frac{(k-2)}{-6} \times \frac{3}{-4} = -1\) OE | m1 | product of grads = −1 in terms of \(k\) |
| \(\Rightarrow k = 10\) | A1 | 3 marks total |
| (d) \(3x - 4y + 5 = 0\) and \(2x - 5y = 6\) | M1 | must use "correct" pair of equations and attempt to eliminate \(y\) (or \(x\)) (generous) |
| \(\Rightarrow Px = \emptyset\) or \(Ry = 5\) | A1 | |
| \(x = -7\), \(y = -4\) | A1 | 3 marks total |
| \((-7, -4)\) |
**(a)** $3p - 4(p + 2) + 5 = 0$ | M1 | condone omission of brackets or one sign error
$\Rightarrow p = -3$ | A1 | 2 marks total
**(b)** $y = \frac{3}{4}x + \frac{5}{4}$ | M1 | rearranging into form $y = \pm\frac{3}{4}x + c$
(gradient $AB =$ ) $\frac{3}{4}$ | A1 | condone slips in rearranging if gradient is correct; 2 marks total
**(c)** (gradient $AC =$ ) $\frac{k-2}{-5-1}$ or $\frac{2-k}{1-(-5)}$ | M1 | (condone one sign error)
"their" $\frac{(k-2)}{-6} \times \frac{3}{-4} = -1$ OE | m1 | product of grads = −1 in terms of $k$
$\Rightarrow k = 10$ | A1 | 3 marks total
**(d)** $3x - 4y + 5 = 0$ and $2x - 5y = 6$ | M1 | must use "correct" pair of equations and attempt to eliminate $y$ (or $x$) (generous)
$\Rightarrow Px = \emptyset$ or $Ry = 5$ | A1 |
$x = -7$, $y = -4$ | A1 | 3 marks total
| | | $(-7, -4)$
**Total: 10 marks**
---1 The line $A B$ has equation $3 x - 4 y + 5 = 0$.
\begin{enumerate}[label=(\alph*)]
\item The point with coordinates $( p , p + 2 )$ lies on the line $A B$. Find the value of the constant $p$.
\item Find the gradient of $A B$.
\item The point $A$ has coordinates ( 1,2 ). The point $C ( - 5 , k )$ is such that $A C$ is perpendicular to $A B$. Find the value of $k$.
\item The line $A B$ intersects the line with equation $2 x - 5 y = 6$ at the point $D$. Find the coordinates of $D$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2013 Q1 [10]}}