| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 13 |
| 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 multi-part coordinate geometry question testing standard techniques: distance formula, perpendicular gradients, equation of a line, and triangle area. All parts follow routine procedures with no problem-solving insight required, making it easier than average but not trivial due to the multiple steps involved. |
| 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 | Marks | Guidance |
| \(QR=\sqrt{(7-1)^2+(0-3)^2}\) | M1 | Attempt \(QR\) or \(QR^2\); implied by \(6^2+3^2\) |
| \(=\sqrt{36+9}\) or \(\sqrt{45}\) | A1 | Condone \(\pm\) |
| \(=3\sqrt{5}\) or \(a=3\) | A1 | \(\pm3\sqrt{5}\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient of \(QR\): \(\frac{3-0}{1-7}=\frac{3}{-6}=-\frac{1}{2}\) | M1, A1 | |
| Gradient of \(l_2\) is \(-\frac{1}{-\frac{1}{2}}=2\) | M1 | Perpendicular rule |
| Equation of \(l_2\): \(y-3=2(x-1)\) or \(\frac{y-3}{x-1}=2\) [i.e. \(y=2x+1\)] | M1 A1ft | Requires all 3 Ms; ft their gradient of \(QR\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P\) is \((0,1)\) | B1 | Allow "\(x=0\), \(y=1\)" if clearly identifiable as \(P\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(PQ=\sqrt{(1-x_P)^2+(3-y_P)^2}=\sqrt{1^2+2^2}=\sqrt{5}\) | M1, A1 | Follow through coordinates of \(P\) |
| Area \(=\frac{1}{2}QR\times PQ=\frac{1}{2}\times3\sqrt{5}\times\sqrt{5}=\frac{15}{2}\) or \(7.5\) | dM1, A1 | Dependent on first M; some working must be seen |
## Question 10:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $QR=\sqrt{(7-1)^2+(0-3)^2}$ | M1 | Attempt $QR$ or $QR^2$; implied by $6^2+3^2$ |
| $=\sqrt{36+9}$ or $\sqrt{45}$ | A1 | Condone $\pm$ |
| $=3\sqrt{5}$ or $a=3$ | A1 | $\pm3\sqrt{5}$ is A0 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of $QR$: $\frac{3-0}{1-7}=\frac{3}{-6}=-\frac{1}{2}$ | M1, A1 | |
| Gradient of $l_2$ is $-\frac{1}{-\frac{1}{2}}=2$ | M1 | Perpendicular rule |
| Equation of $l_2$: $y-3=2(x-1)$ or $\frac{y-3}{x-1}=2$ [i.e. $y=2x+1$] | M1 A1ft | Requires all 3 Ms; ft their gradient of $QR$ |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P$ is $(0,1)$ | B1 | Allow "$x=0$, $y=1$" if clearly identifiable as $P$ |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $PQ=\sqrt{(1-x_P)^2+(3-y_P)^2}=\sqrt{1^2+2^2}=\sqrt{5}$ | M1, A1 | Follow through coordinates of $P$ |
| Area $=\frac{1}{2}QR\times PQ=\frac{1}{2}\times3\sqrt{5}\times\sqrt{5}=\frac{15}{2}$ or $7.5$ | dM1, A1 | Dependent on first M; some working must be seen |10.
\begin{tikzpicture}[>=latex]
% Axes
\draw[->] (-0.8, 0) -- (4.5, 0) node[below] {$x$};
\draw[->] (0, -0.5) -- (0, 3.5) node[left] {$y$};
\node at (-0.1, -0.3) {$O$};
% Line l1
\draw (-0.2, 1.7) -- (4.1, -0.1) node[pos=0.45, above right] {$l_1$};
% Line l2
\draw (-0.4, -0.5) -- (1.1, 3.1) node[pos=0.9, right] {$l_2$};
% Intersection points and labels
\node at (-0.15, 0.45) {$P$};
\node at (0.35, 1.65) {$Q$};
\node at (3.8, -0.3) {$R$};
\end{tikzpicture}
The points $Q ( 1,3 )$ and $R ( 7,0 )$ lie on the line $l _ { 1 }$, as shown in Figure 2.\\
The length of $Q R$ is $a \sqrt { } 5$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$.
The line $l _ { 2 }$ is perpendicular to $l _ { 1 }$, passes through $Q$ and crosses the $y$-axis at the point $P$, as shown in Figure 2.
Find
\item an equation for $l _ { 2 }$,
\item the coordinates of $P$,
\item the area of $\triangle P Q R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2008 Q10 [13]}}