| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Coordinates from geometric constraints |
| Difficulty | Moderate -0.8 This is a straightforward multi-part coordinate geometry question testing basic skills: gradient formula, equation of a line, midpoint formula, and distance formula. All parts are routine calculations with no problem-solving insight required. Part (c) involves solving a simple quadratic equation. This is easier than average A-level content, being typical of early Core 1/2 material. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m = \frac{5-1}{-1-4}\) | M1 | Attempts \(\frac{\text{change in }y}{\text{change in }x}\). Condone one sign slip. Maybe implied by \(\pm\frac{4}{5}\) |
| \(m = -\frac{4}{5}\) | A1 | cao. Correct answer only scores both marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5 = -m + c\), \(1 = 4m + c \Rightarrow 5-1 = -m-4m \Rightarrow m = \ldots\) | M1 | Correct method for the gradient |
| \(m = -\frac{4}{5}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y - 5 = -\frac{4}{5}(x+1)\) or \(y - 1 = -\frac{4}{5}(x-4)\) | M1 | Uses \(A\) or \(B\) and their \(m\) in a correct straight line method. If using \(y=mx+c\) must reach as far as \(c = \ldots\). Attempting the normal is M0 |
| \(4x + 5y - 21 = 0\) | A1 | Allow any integer multiple |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M\) is \(\left(\frac{3}{2}, 3\right)\) | B1 | Correct midpoint |
| \(MC^2 = \left(5 - \frac{3}{2}\right)^2 + (k-3)^2\) | M1 | Correct use of Pythagoras for \(MC\). E.g. sight of \(\left(5-\frac{3}{2}\right)^2 + h^2\) or \(\sqrt{\left(5-\frac{3}{2}\right)^2 + h^2}\) where \(h = k-3\) or \(h=k\) |
| \(\left(5 - \frac{3}{2}\right)^2 + (k-3)^2 = 13 \Rightarrow k = \ldots\) | dM1 | Uses \(\sqrt{13}\) correctly to find value of \(k\). Must be correct method. \(\left(5-\frac{3}{2}\right)^2 + (k-3)^2 = 13^2\) scores M0. Dependent on first M mark |
| \(k = 3 \pm \frac{\sqrt{3}}{2}\) | A1 | Both. Accept e.g. \(\frac{24\pm\sqrt{48}}{8}\), \(\frac{6\pm\sqrt{3}}{2}\). Ignore how they are referenced, e.g. no need for \(k=\ldots\) |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m = \frac{5-1}{-1-4}$ | M1 | Attempts $\frac{\text{change in }y}{\text{change in }x}$. Condone one sign slip. Maybe implied by $\pm\frac{4}{5}$ |
| $m = -\frac{4}{5}$ | A1 | cao. Correct answer only scores both marks |
**Way 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5 = -m + c$, $1 = 4m + c \Rightarrow 5-1 = -m-4m \Rightarrow m = \ldots$ | M1 | Correct method for the gradient |
| $m = -\frac{4}{5}$ | A1 | cao |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y - 5 = -\frac{4}{5}(x+1)$ or $y - 1 = -\frac{4}{5}(x-4)$ | M1 | Uses $A$ or $B$ and their $m$ in a correct straight line method. If using $y=mx+c$ must reach as far as $c = \ldots$. **Attempting the normal is M0** |
| $4x + 5y - 21 = 0$ | A1 | Allow any integer multiple |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M$ is $\left(\frac{3}{2}, 3\right)$ | B1 | Correct midpoint |
| $MC^2 = \left(5 - \frac{3}{2}\right)^2 + (k-3)^2$ | M1 | Correct use of Pythagoras for $MC$. E.g. sight of $\left(5-\frac{3}{2}\right)^2 + h^2$ or $\sqrt{\left(5-\frac{3}{2}\right)^2 + h^2}$ where $h = k-3$ or $h=k$ |
| $\left(5 - \frac{3}{2}\right)^2 + (k-3)^2 = 13 \Rightarrow k = \ldots$ | dM1 | Uses $\sqrt{13}$ correctly to find value of $k$. Must be correct method. $\left(5-\frac{3}{2}\right)^2 + (k-3)^2 = 13^2$ scores M0. **Dependent on first M mark** |
| $k = 3 \pm \frac{\sqrt{3}}{2}$ | A1 | Both. Accept e.g. $\frac{24\pm\sqrt{48}}{8}$, $\frac{6\pm\sqrt{3}}{2}$. Ignore how they are referenced, e.g. no need for $k=\ldots$ |
---7. The point $A$ has coordinates $( - 1,5 )$ and the point $B$ has coordinates $( 4,1 )$.
The line $l$ passes through the points $A$ and $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of $l$.
\item Find an equation for $l$, giving your answer in the form $a x + b y + c = 0$ where $a , b$ and $c$ are integers.
The point $M$ is the midpoint of $A B$.
The point $C$ has coordinates $( 5 , k )$ where $k$ is a constant.\\
Given that the distance from $M$ to $C$ is $\sqrt { 13 }$
\item find the exact possible values of the constant $k$.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q7 [8]}}