| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Parallel line through point |
| Difficulty | Standard +0.3 This is a standard C1 coordinate geometry question requiring finding a line equation from two points, using the parallel lines distance formula, and calculating area. While multi-part, each step uses routine techniques (gradient, perpendicular distance formula, area = base × height) with no novel problem-solving required. Slightly above average due to the distance formula application and multi-step nature, but still straightforward for C1 level. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.10a Vectors in 2D: i,j notation and column vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (a) grad \(= \frac{4-3}{3-(-1)} = \frac{1}{4}\) | M1 A1 | |
| \(\therefore y - 3 = \frac{1}{4}(x+1)\) | M1 | |
| \(4y - 12 = x + 1\) | ||
| \(x - 4y + 13 = 0\) | A1 | |
| (b) perp grad \(= -\frac{1}{\frac{1}{4}} = -4\) | M1 | |
| line through A, perp \(l_1\): \(y - 3 = -4(x+1)\) | M1 | |
| \(y = -4x - 1\) | A1 | |
| intersection with \(l_2\): \(x - 4(-4x-1) - 21 = 0\) | ||
| \(x = 1, \therefore (1, -5)\) | M1 A1 | |
| dist. A to \((1, -5) = \sqrt{(1+1)^2 + (-5-3)^2} = \sqrt{4 + 64} = \sqrt{68}\) | M1 | |
| \(\therefore\) dist. between lines \(= \sqrt{68} = \sqrt{4 \times 17} = 2\sqrt{17}\) | A1 | \([k=2]\) |
| (c) \(AB = \sqrt{(3+1)^2 + (4-3)^2} = \sqrt{16 + 1} = \sqrt{17}\) | M1 | |
| area \(= \sqrt{17} \times 2\sqrt{17} = 34\) | A1 | (13) |
(a) grad $= \frac{4-3}{3-(-1)} = \frac{1}{4}$ | M1 A1 |
$\therefore y - 3 = \frac{1}{4}(x+1)$ | M1 |
$4y - 12 = x + 1$ | |
$x - 4y + 13 = 0$ | A1 |
(b) perp grad $= -\frac{1}{\frac{1}{4}} = -4$ | M1 |
line through A, perp $l_1$: $y - 3 = -4(x+1)$ | M1 |
$y = -4x - 1$ | A1 |
intersection with $l_2$: $x - 4(-4x-1) - 21 = 0$ | |
$x = 1, \therefore (1, -5)$ | M1 A1 |
dist. A to $(1, -5) = \sqrt{(1+1)^2 + (-5-3)^2} = \sqrt{4 + 64} = \sqrt{68}$ | M1 |
$\therefore$ dist. between lines $= \sqrt{68} = \sqrt{4 \times 17} = 2\sqrt{17}$ | A1 | $[k=2]$ |
(c) $AB = \sqrt{(3+1)^2 + (4-3)^2} = \sqrt{16 + 1} = \sqrt{17}$ | M1 |
area $= \sqrt{17} \times 2\sqrt{17} = 34$ | A1 | (13)10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5e8f154b-c232-49ee-a798-f61ff08ca0b9-4_663_1113_950_402}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the parallelogram $A B C D$.\\
The points $A$ and $B$ have coordinates $( - 1,3 )$ and $( 3,4 )$ respectively and lie on the straight line $l _ { 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for $l _ { 1 }$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
The points $C$ and $D$ lie on the straight line $l _ { 2 }$ which has the equation $x - 4 y - 21 = 0$.
\item Show that the distance between $l _ { 1 }$ and $l _ { 2 }$ is $k \sqrt { 17 }$, where $k$ is an integer to be found.
\item Find the area of parallelogram $A B C D$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q10 [13]}}