| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Line intersecting general conic |
| Difficulty | Standard +0.3 This is a straightforward C1 question involving substitution to find intersection points, followed by verifying perpendicularity using gradients. While it requires multiple steps (finding A, solving simultaneous equations, calculating gradients), each step uses standard techniques with no novel insight needed. Slightly above average difficulty due to the algebraic manipulation and the perpendicularity verification, but well within typical C1 scope. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution |
| Answer | Marks | Guidance |
|---|---|---|
| \((2, 0)\) (or \(x = 2\), \(y = 0\)) | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y^2 = 4\left(\frac{3y+12}{2} - 2\right)\) or \(\left(\frac{2x-12}{3}\right)^2 = 4(x-2)\) | M1 | |
| \(y^2 - 6y - 16 = 0\) or \(x^2 - 21x + 54 = 0\) (or equiv. 3 terms) | A1 | |
| \((y+2)(y-8) = 0\), \(y = \ldots\) or \((x-3)(x-18) = 0\), \(x = \ldots\) | M1 | |
| \(y = -2\), \(y = 8\) or \(x = 3\), \(x = 18\) | A1 | |
| \(x = 3\), \(x = 18\) or \(y = -2\), \(y = 8\) (attempt one for M mark) | M1 A1 ft | A1ft requires both values (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Grad. of \(AQ = \frac{8-0}{18-2}\), Grad. of \(AP = \frac{0-(-2)}{2-3}\) (attempt one for M mark) | M1 A1 ft | |
| \(m_1 \times m_2 = \frac{1}{2} \times -2 = -1\), so \(\angle PAQ\) is a right angle | M1 A1 | A1 is c.s.o. (4 marks) |
| Alternative: Pythagoras: \(AQ = \sqrt{320}\), \(AP = \sqrt{5}\), \(PQ = \sqrt{325}\) | [M1][A1ft] | |
| \(AQ^2 + AP^2 = PQ^2\), so \(\angle PAQ\) is a right angle | [M1, A1] | Requires correct exact working + conclusion |
## Question 7:
### Part (a)
$(2, 0)$ (or $x = 2$, $y = 0$) | B1 | (1 mark)
### Part (b)
$y^2 = 4\left(\frac{3y+12}{2} - 2\right)$ or $\left(\frac{2x-12}{3}\right)^2 = 4(x-2)$ | M1 |
$y^2 - 6y - 16 = 0$ or $x^2 - 21x + 54 = 0$ (or equiv. 3 terms) | A1 |
$(y+2)(y-8) = 0$, $y = \ldots$ or $(x-3)(x-18) = 0$, $x = \ldots$ | M1 |
$y = -2$, $y = 8$ or $x = 3$, $x = 18$ | A1 |
$x = 3$, $x = 18$ or $y = -2$, $y = 8$ (attempt one for M mark) | M1 A1 ft | A1ft requires both values (6 marks)
### Part (c)
Grad. of $AQ = \frac{8-0}{18-2}$, Grad. of $AP = \frac{0-(-2)}{2-3}$ (attempt one for M mark) | M1 A1 ft |
$m_1 \times m_2 = \frac{1}{2} \times -2 = -1$, so $\angle PAQ$ is a right angle | M1 A1 | A1 is c.s.o. (4 marks)
**Alternative:** Pythagoras: $AQ = \sqrt{320}$, $AP = \sqrt{5}$, $PQ = \sqrt{325}$ | [M1][A1ft] |
$AQ^2 + AP^2 = PQ^2$, so $\angle PAQ$ is a right angle | [M1, A1] | Requires correct exact working + conclusion
---7.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{813612f1-92c8-456d-84a2-aa6bb91b8a6a-3_689_1077_927_484}
\end{center}
\end{figure}
Fig. 1 shows the curve with equation $y ^ { 2 } = 4 ( x - 2 )$ and the line with equation $2 x - 3 y = 12$.\\
The curve crosses the $x$-axis at the point $A$, and the line intersects the curve at the points $P$ and $Q$.
\begin{enumerate}[label=(\alph*)]
\item Write down the coordinates of $A$.
\item Find, using algebra, the coordinates of $P$ and $Q$.
\item Show that $\angle P A Q$ is a right angle.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q7 [11]}}