| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Intersection of two lines |
| Difficulty | Moderate -0.8 This is a straightforward C1 coordinate geometry question testing standard techniques: rearranging to find gradient, using perpendicular gradient formula (negative reciprocal), and solving simultaneous equations. All parts are routine textbook exercises requiring only direct application of learned methods with no problem-solving insight needed. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| \(5y = -3x + 7\), gradient \(= -\frac{3}{5}\) | M1, A1 | M1 for rearranging to \(y = mx + c\) form or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Perpendicular gradient \(= \frac{5}{3}\) | B1 | Negative reciprocal of their gradient from (a) |
| \(y - (-3) = \frac{5}{3}(x - (-2))\) | M1 | Using point \((-2, -3)\) with perpendicular gradient |
| \(5x - 3y + 1 = 0\) | A1 | Must be in form \(px + qy + r = 0\) with integers |
| Answer | Marks | Guidance |
|---|---|---|
| Solving simultaneously: \(3x + 5y = 7\) and \(2x - 3y = 30\) | M1 | Attempt to eliminate one variable |
| \(9x + 15y = 21\) and \(10x - 15y = 150\), so \(19x = 171\) | M1 | Correct elimination method |
| \(x = 9\), \(y = -\frac{16}{5}\), so \(A = \left(9, -\frac{16}{5}\right)\) | A1 | Both coordinates correct |
# Question 1:
## Part (a)
| $5y = -3x + 7$, gradient $= -\frac{3}{5}$ | M1, A1 | M1 for rearranging to $y = mx + c$ form or equivalent |
## Part (b)
| Perpendicular gradient $= \frac{5}{3}$ | B1 | Negative reciprocal of their gradient from (a) |
| $y - (-3) = \frac{5}{3}(x - (-2))$ | M1 | Using point $(-2, -3)$ with perpendicular gradient |
| $5x - 3y + 1 = 0$ | A1 | Must be in form $px + qy + r = 0$ with integers |
## Part (c)
| Solving simultaneously: $3x + 5y = 7$ and $2x - 3y = 30$ | M1 | Attempt to eliminate one variable |
| $9x + 15y = 21$ and $10x - 15y = 150$, so $19x = 171$ | M1 | Correct elimination method |
| $x = 9$, $y = -\frac{16}{5}$, so $A = \left(9, -\frac{16}{5}\right)$ | A1 | Both coordinates correct |
---1 The line $A B$ has equation $3 x + 5 y = 7$.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of $A B$.
\item Find an equation of the line that is perpendicular to the line $A B$ and which passes through the point $( - 2 , - 3 )$. Express your answer in the form $p x + q y + r = 0$, where $p , q$ and $r$ are integers.
\item The line $A C$ has equation $2 x - 3 y = 30$. Find the coordinates of $A$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2015 Q1 [8]}}