| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Coordinates from geometric constraints |
| Difficulty | Standard +0.3 Part (a) is a routine gradient-point form to general form conversion. Part (b) requires using perpendicular gradients (product = -1) to find an unknown coordinate, which is standard C1/C2 material but involves an extra step of algebraic manipulation. The question is slightly above average due to the two-part structure and the geometric constraint application, but remains a textbook exercise with no novel insight required. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient \(PQ = \frac{y_1 - y_2}{x_1 - x_2} = \frac{7-4}{4-(-1)} = \frac{3}{5}\) | M1A1 | M1 for attempt at gradient using both \((4,7)\) and \((-1,4)\); accept \(\frac{7-4}{4-(-1)}\) as evidence. Allow one sign error. Do NOT accept \(\frac{7+4}{4-1}\) or \(\frac{4--1}{7-4}\) |
| Equation of line \(PQ\): \(\frac{3}{5} = \frac{y-7}{x-4}\), \(\frac{3}{5} = \frac{y-4}{x--1}\) oe | M1 | Attempts equation of line using their gradient and a point. Accept \(y-7 = \text{their}\frac{3}{5}(x-4)\) with both signs correct |
| \(\pm k(3x - 5y + 23 = 0)\), \(k\) an integer | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses gradient \(PR = -\frac{5}{3} \Rightarrow \frac{-7}{p+1} = -\frac{5}{3}\) | M1 | Attempts to use perpendicular gradients to set up equation in \(p\). Accept \(\frac{3}{5} \times \frac{4--3}{-1-p} = -1\). Also accept Pythagoras: \(PQ^2 + PR^2 = QR^2 \Rightarrow (p+1)^2 + 7^2 + 5^2 + 3^2 = (p-4)^2 + 10^2\) |
| \(5p + 5 = 21 \Rightarrow p = \frac{16}{5}\) oe | dM1, A1 | dM1 dependent on previous M; scored for proceeding to and solving a linear equation in \(p\). A1 accept \(3.2\) or \(3\frac{1}{5}\), also accept written as \(x =\) |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient $PQ = \frac{y_1 - y_2}{x_1 - x_2} = \frac{7-4}{4-(-1)} = \frac{3}{5}$ | M1A1 | M1 for attempt at gradient using both $(4,7)$ and $(-1,4)$; accept $\frac{7-4}{4-(-1)}$ as evidence. Allow one sign error. Do NOT accept $\frac{7+4}{4-1}$ or $\frac{4--1}{7-4}$ |
| Equation of line $PQ$: $\frac{3}{5} = \frac{y-7}{x-4}$, $\frac{3}{5} = \frac{y-4}{x--1}$ oe | M1 | Attempts equation of line using their gradient and a point. Accept $y-7 = \text{their}\frac{3}{5}(x-4)$ with both signs correct |
| $\pm k(3x - 5y + 23 = 0)$, $k$ an integer | A1 | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses gradient $PR = -\frac{5}{3} \Rightarrow \frac{-7}{p+1} = -\frac{5}{3}$ | M1 | Attempts to use perpendicular gradients to set up equation in $p$. Accept $\frac{3}{5} \times \frac{4--3}{-1-p} = -1$. Also accept Pythagoras: $PQ^2 + PR^2 = QR^2 \Rightarrow (p+1)^2 + 7^2 + 5^2 + 3^2 = (p-4)^2 + 10^2$ |
| $5p + 5 = 21 \Rightarrow p = \frac{16}{5}$ oe | dM1, A1 | dM1 dependent on previous M; scored for proceeding to and solving a linear equation in $p$. A1 accept $3.2$ or $3\frac{1}{5}$, also accept written as $x =$ |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b85872d4-00b2-499b-9765-f7559d3de66a-07_953_929_219_422}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Diagram not drawn to scale
Figure 3 shows the points $P , Q$ and $R$.
Points $P$ and $Q$ have coordinates ( $- 1,4$ ) and ( 4,7 ) respectively.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for the straight line passing through points $P$ and $Q$.
Give your answer in the form $a x + b y + c = 0$, where $a$, $b$ and $c$ are integers.
The point $R$ has coordinates ( $p , - 3$ ), where $p$ is a positive constant.
Given that angle $Q P R = 90 ^ { \circ }$,
\item find the value of $p$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2014 Q5 [7]}}