| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular bisector of segment |
| Difficulty | Moderate -0.8 This is a straightforward multi-part coordinate geometry question requiring only standard techniques: substituting points into a line equation, using the distance formula, finding midpoints, and using the perpendicular gradient relationship. All steps are routine applications of formulas with no problem-solving insight required, making it easier than average but not trivial due to the multiple parts. |
| 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 |
| \(p = 4\) or \(q = 5\) | B1 | One correct value. May be implied by e.g. when \(x=-1, y=4\) or when \(y=2, x=5\) |
| \(p = 4\) and \(q = 5\) | B1 | Both correct values |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AB^2 = (\text{"4"}-2)^2 + (-1-\text{"5"})^2\) or \(AB = \sqrt{(\text{"4"}-2)^2+(-1-\text{"5"})^2}\) | M1 | Correct Pythagoras method using \((-1, \text{"4"})\) and \((\text{"5"}, 2)\) to find \(AB\) or \(AB^2\) |
| \((AB) = 2\sqrt{10}\) | A1 | \(2\sqrt{10}\) only |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M = \left(\frac{-1+\text{"5"}}{2}, \frac{\text{"4"}+2}{2}\right) = (2, 3)\) | M1 | Correct midpoint method. May be implied by at least one correct coordinate |
| Gradient of \(l_1 = -\frac{1}{3}\) | B1 | Correct gradient of \(l_1\). May be implied by a correct perpendicular gradient |
| Perpendicular gradient \(= 3\) | M1 | Correct perpendicular gradient rule, e.g. \(m = \frac{-1}{-\frac{1}{3}}\) or \(\frac{-1}{3} \times m = -1 \Rightarrow m = ...\) |
| \(y - \text{"3"} = \text{"3"}(x - \text{"2"})\) or \(y = mx + x \Rightarrow \text{"3"} = \text{"3"} \times \text{"2"} + c \Rightarrow c = ...\) | M1 | Correct straight line method using their midpoint and a "changed" gradient |
| \(y = 3x - 3\) | A1 | cao |
| Alternative last 4 marks: \(3x - y + c = 0\) | B1M1 | B1: "\(3x-y\)"; M1: \(3x - y + c = 0\) |
| \(3(2) - 3 + c = 0 \Rightarrow c = -3\) | M1 | Correct method to find \(c\) using their values |
| \(y = 3x - 3\) | A1 | cao |
| Total: 5 marks |
## Question 12:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $p = 4$ or $q = 5$ | B1 | One correct value. May be implied by e.g. when $x=-1, y=4$ or when $y=2, x=5$ |
| $p = 4$ and $q = 5$ | B1 | Both correct values |
| **Total: 2 marks** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB^2 = (\text{"4"}-2)^2 + (-1-\text{"5"})^2$ or $AB = \sqrt{(\text{"4"}-2)^2+(-1-\text{"5"})^2}$ | M1 | Correct Pythagoras method using $(-1, \text{"4"})$ and $(\text{"5"}, 2)$ to find $AB$ or $AB^2$ |
| $(AB) = 2\sqrt{10}$ | A1 | $2\sqrt{10}$ only |
| **Total: 2 marks** | | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M = \left(\frac{-1+\text{"5"}}{2}, \frac{\text{"4"}+2}{2}\right) = (2, 3)$ | M1 | Correct midpoint method. May be implied by at least one correct coordinate |
| Gradient of $l_1 = -\frac{1}{3}$ | B1 | Correct gradient of $l_1$. May be implied by a correct perpendicular gradient |
| Perpendicular gradient $= 3$ | M1 | Correct perpendicular gradient rule, e.g. $m = \frac{-1}{-\frac{1}{3}}$ or $\frac{-1}{3} \times m = -1 \Rightarrow m = ...$ |
| $y - \text{"3"} = \text{"3"}(x - \text{"2"})$ or $y = mx + x \Rightarrow \text{"3"} = \text{"3"} \times \text{"2"} + c \Rightarrow c = ...$ | M1 | Correct straight line method using their midpoint and a "changed" gradient |
| $y = 3x - 3$ | A1 | cao |
| **Alternative last 4 marks:** $3x - y + c = 0$ | B1M1 | B1: "$3x-y$"; M1: $3x - y + c = 0$ |
| $3(2) - 3 + c = 0 \Rightarrow c = -3$ | M1 | Correct method to find $c$ using their values |
| $y = 3x - 3$ | A1 | cao |
| **Total: 5 marks** | | |
---\begin{enumerate}
\item The line $l _ { 1 }$ has equation $x + 3 y - 11 = 0$
\end{enumerate}
The point $A$ and the point $B$ lie on $l _ { 1 }$\\
Given that $A$ has coordinates ( $- 1 , p$ ) and $B$ has coordinates ( $q , 2$ ), where $p$ and $q$ are integers,\\
(a) find the value of $p$ and the value of $q$,\\
(b) find the length of $A B$, giving your answer as a simplified surd.
The line $l _ { 2 }$ is perpendicular to $l _ { 1 }$ and passes through the midpoint of $A B$.\\
(c) Find an equation for $l _ { 2 }$ giving your answer in the form $y = m x + c$, where $m$ and $c$ are constants to be found.
\hfill \mbox{\textit{Edexcel C12 2018 Q12 [9]}}