| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular from point to line |
| Difficulty | Moderate -0.3 This is a standard C1 coordinate geometry question testing routine skills: finding intercepts, parallel/perpendicular lines, and triangle area. While multi-part with 13 marks total, each step uses basic techniques (substitution, gradient rules, simultaneous equations) with no novel problem-solving required. Slightly easier than average due to straightforward application of standard methods, though the perpendicular line calculation in part (d) adds minor complexity. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.10g Problem solving with vectors: in geometry |
| Answer | Marks |
|---|---|
| (a) \(y = 0 \therefore x = 7 \Rightarrow A(7, 0)\) | M1 A1 |
| (b) \(l_1: y = 14 - 2x \therefore\) grad \(= -2\) | B1 |
| \(l_2: y - 6 = -2(x + 6)\) | M1 |
| \(y = -2x - 6\) | A1 |
| (c) \(y = 0 \therefore x = -3 \Rightarrow C(-3, 0)\) | B1 |
| (d) grad \(CD = \frac{-1}{-2} = \frac{1}{2}\) | M1 |
| eqn \(CD: y - 0 = \frac{1}{2}(x + 3)\) | M1 A1 |
| intersection with \(l_1: \frac{1}{2}(x + 3) = 14 - 2x\) | |
| \(x = 5\) | M1 |
| \(y = 14 - (2 \times 5) = 4\) | |
| \(\therefore D(5, 4)\) | A1 |
| (e) \(AC = 7 - (-3) = 10\) | |
| area \(= \frac{1}{2} \times 10 \times 4 = 20\) | M1 A1 |
**(a)** $y = 0 \therefore x = 7 \Rightarrow A(7, 0)$ | M1 A1 |
**(b)** $l_1: y = 14 - 2x \therefore$ grad $= -2$ | B1 |
$l_2: y - 6 = -2(x + 6)$ | M1 |
$y = -2x - 6$ | A1 |
**(c)** $y = 0 \therefore x = -3 \Rightarrow C(-3, 0)$ | B1 |
**(d)** grad $CD = \frac{-1}{-2} = \frac{1}{2}$ | M1 |
eqn $CD: y - 0 = \frac{1}{2}(x + 3)$ | M1 A1 |
intersection with $l_1: \frac{1}{2}(x + 3) = 14 - 2x$ |
$x = 5$ | M1 |
$y = 14 - (2 \times 5) = 4$ |
$\therefore D(5, 4)$ | A1 |
**(e)** $AC = 7 - (-3) = 10$ |
area $= \frac{1}{2} \times 10 \times 4 = 20$ | M1 A1 |
**Total: 13 marks**The straight line $l_1$ has equation $2x + y - 14 = 0$ and crosses the $x$-axis at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $A$. [2]
\end{enumerate}
The straight line $l_2$ is parallel to $l_1$ and passes through the point $B(-6, 6)$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find an equation for $l_2$ in the form $y = mx + c$. [3]
\end{enumerate}
The line $l_2$ crosses the $x$-axis at the point $C$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the coordinates of $C$. [1]
\end{enumerate}
The point $D$ lies on $l_1$ and is such that $CD$ is perpendicular to $l_1$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Show that $D$ has coordinates $(5, 4)$. [5]
\item Find the area of triangle $ACD$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q10 [13]}}