| Exam Board | OCR |
|---|---|
| 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 multi-part coordinate geometry question requiring routine techniques: finding intercepts, parallel line equations, perpendicular lines, and triangle area. While it has 5 parts totaling 13 marks, each step uses straightforward methods with no novel insight required. The 'show that' in part (iv) adds slight rigor but the overall difficulty remains slightly below average for A-level. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks |
|---|---|
| \(y = 0\) ∴ \(x = 7\) ⟹ \(A(7, 0)\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(l_1: y = 14 - 2x\) ∴ grad \(= -2\) | B1 |
| \(l_2: y - 6 = -2(x + 6)\) | M1 |
| \(y = -2x - 6\) | A1 |
| Answer | Marks |
|---|---|
| \(y = 0\) ∴ \(x = -3\) ⟹ \(C(-3, 0)\) | B1 |
| Answer | Marks |
|---|---|
| grad \(CD = \frac{-1}{-2} = \frac{1}{2}\) | M1 |
| eqn \(CD: y - 0 = \frac{1}{2}(x + 3)\) | M1 A1 |
| intersection with \(l_1: \frac{1}{2}(x + 3) = 14 - 2x\) | M1 |
| \(x = 5\) | M1 |
| \(y = 14 - (2 \times 5) = 4\) ∴ \(D(5, 4)\) | A1 |
| Answer | Marks |
|---|---|
| \(AC = 7 - (-3) = 10\) | M1 A1 |
| area \(= \frac{1}{2} \times 10 \times 4 = 20\) | M1 A1 |
| (13) |
| Answer | Marks |
|---|---|
| Total | (72) |
## (i)
$y = 0$ ∴ $x = 7$ ⟹ $A(7, 0)$ | M1 A1
## (ii)
$l_1: y = 14 - 2x$ ∴ grad $= -2$ | B1
$l_2: y - 6 = -2(x + 6)$ | M1
$y = -2x - 6$ | A1
## (iii)
$y = 0$ ∴ $x = -3$ ⟹ $C(-3, 0)$ | B1
## (iv)
grad $CD = \frac{-1}{-2} = \frac{1}{2}$ | M1
eqn $CD: y - 0 = \frac{1}{2}(x + 3)$ | M1 A1
intersection with $l_1: \frac{1}{2}(x + 3) = 14 - 2x$ | M1
$x = 5$ | M1
$y = 14 - (2 \times 5) = 4$ ∴ $D(5, 4)$ | A1
## (v)
$AC = 7 - (-3) = 10$ | M1 A1
area $= \frac{1}{2} \times 10 \times 4 = 20$ | M1 A1
| (13)
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**Total** | (72)The straight line $l_1$ has equation $2x + y - 14 = 0$ and crosses the $x$-axis at the point $A$.
\begin{enumerate}[label=(\roman*)]
\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=(\roman*)]
\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=(\roman*)]
\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=(\roman*)]
\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{OCR C1 Q10 [13]}}