| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Moderate -0.8 This is a straightforward C1 coordinate geometry question requiring basic distance formula, gradient calculation, perpendicular gradient relationship, and simple algebraic manipulation. All techniques are standard textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((OA^2 =) 6^2 + (-4)^2\) ; \((OB^2 =) (-2)^2 + 7^2\) | M1 | Either correct PI by 52 or 53 seen |
| \((OA^2 =) 52\) and \((OB^2 =) 53\) or \((OA =)\sqrt{52}\) and \((OB =)\sqrt{53}\) | A1 | Both correct values 52 or \(\sqrt{52}\) and 53 or \(\sqrt{53}\) seen |
| \(OA = \sqrt{52}\) and \(OB = \sqrt{53}\) \(\Rightarrow OA < OB\) | A1 | Correct working + concluding statement involving \(OA\) and/or \(OB\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\text{grad } AB = \frac{7+4}{-2-6}\) | M1 | Condone one sign error |
| \(= -\frac{11}{8}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y - (-4) = \text{their grad } AB'(x-6)\) or \(y - 7 = \text{their grad } AB'(x-(-2))\) | M1 | Or \(y = \text{their grad } AB' \cdot x + c\) and attempt to find \(c\) using \(x=6, y=-4\) or \(x=-2, y=7\) |
| \(y + 4 = -\frac{11}{8}(x-6)\) OE | A1 | Any correct form e.g. \(y = -\frac{11}{8}x + \frac{34}{8}\); must simplify \(--\) to \(+\) |
| \(\Rightarrow 11x + 8y = 34\) | A1 | Condone \(8y + 11x = 34\) or any multiple |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((\text{grad } AC =)\ \frac{8}{11}\) | B1\(\checkmark\) | FT \(-1/\text{their grad } AB\) |
| \(\frac{4}{k-6} = \text{'their } \frac{8}{11}\text{'}\) OE | M1 | Equating gradients; LHS must be correct and RHS is "attempt" at perp grad to \(AB\) |
| \(\Rightarrow 2k - 12 = 11\) | ||
| \(\Rightarrow k = \frac{23}{2}\) | A1cso | \(k = 11.5\) OE |
# Question 1:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $(OA^2 =) 6^2 + (-4)^2$ ; $(OB^2 =) (-2)^2 + 7^2$ | M1 | Either correct PI by 52 or 53 seen |
| $(OA^2 =) 52$ **and** $(OB^2 =) 53$ or $(OA =)\sqrt{52}$ **and** $(OB =)\sqrt{53}$ | A1 | Both correct values 52 or $\sqrt{52}$ **and** 53 or $\sqrt{53}$ seen |
| $OA = \sqrt{52}$ and $OB = \sqrt{53}$ $\Rightarrow OA < OB$ | A1 | Correct working + concluding statement involving $OA$ and/or $OB$ |
## Part (b)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $\text{grad } AB = \frac{7+4}{-2-6}$ | M1 | Condone one sign error |
| $= -\frac{11}{8}$ | A1 | |
## Part (b)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| $y - (-4) = \text{their grad } AB'(x-6)$ or $y - 7 = \text{their grad } AB'(x-(-2))$ | M1 | Or $y = \text{their grad } AB' \cdot x + c$ and attempt to find $c$ using $x=6, y=-4$ or $x=-2, y=7$ |
| $y + 4 = -\frac{11}{8}(x-6)$ OE | A1 | Any correct form e.g. $y = -\frac{11}{8}x + \frac{34}{8}$; must simplify $--$ to $+$ |
| $\Rightarrow 11x + 8y = 34$ | A1 | Condone $8y + 11x = 34$ or any multiple |
## Part (c)
| Working | Mark | Guidance |
|---------|------|----------|
| $(\text{grad } AC =)\ \frac{8}{11}$ | B1$\checkmark$ | FT $-1/\text{their grad } AB$ |
| $\frac{4}{k-6} = \text{'their } \frac{8}{11}\text{'}$ OE | M1 | Equating gradients; LHS must be correct and RHS is "attempt" at perp grad to $AB$ |
| $\Rightarrow 2k - 12 = 11$ | | |
| $\Rightarrow k = \frac{23}{2}$ | A1cso | $k = 11.5$ OE |
---1 The point $A$ has coordinates (6, -4) and the point $B$ has coordinates (-2, 7).
\begin{enumerate}[label=(\alph*)]
\item Given that the point $O$ has coordinates $( 0,0 )$, show that the length of $O A$ is less than the length of $O B$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the gradient of $A B$.
\item Find an equation of the line $A B$ in the form $p x + q y = r$, where $p , q$ and $r$ are integers.
\end{enumerate}\item The point $C$ has coordinates $( k , 0 )$. The line $A C$ is perpendicular to the line $A B$. Find the value of the constant $k$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2012 Q1 [11]}}