| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Rectangle or parallelogram vertices |
| Difficulty | Moderate -0.3 This is a standard coordinate geometry question requiring routine techniques: finding gradients, perpendicular/parallel line equations, distance formula, and trapezium area. While it has multiple parts (9 marks total), each step follows directly from the previous with no novel insight required. The 'show that' in part (a)(ii) provides the answer, reducing problem-solving demand. Slightly easier than average due to its structured, procedural nature. |
| Spec | 1.02b Surds: manipulation and rationalising denominators1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| 4(a)(i) | Uses coordinates of A and B to | |
| find gradient of AB | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| (any form) | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 4(a)(ii) | Uses perpendicular gradients | |
| property. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| required gradient of -2 | 1.1b | A1F |
| Answer | Marks | Guidance |
|---|---|---|
| both lines | 1.1b | A1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 4(b)(i) | Calculates length of AB and CD. | |
| At least one correct. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains correct simplified sum | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 4(b)(ii) | Calculates AD and applies | |
| trapezium area formula | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains correct area | 1.1b | A1 |
| Subtotal | 2 | |
| Question Total | 9 | |
| Q | Marking instructions | AO |
Question 4:
--- 4(a)(i) ---
4(a)(i) | Uses coordinates of A and B to
find gradient of AB | 3.1a | M1 | Grad AB = = = Grad DC
3 1
Equation is y – 4 =
6 2
1
2y = x + 5
2(𝑥𝑥−3)
Obtains correct equation of CD
(any form) | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 4(a)(ii) ---
4(a)(ii) | Uses perpendicular gradients
property. | 1.1a | M1 | Grad DA = = –2
−1
Grad AB
Equation is y + 2 =
y = –2x
−2(𝑥𝑥−1)
Intersect at (–1, 2) = D
Obtains correct equation of AD
using their gradient (any form)
Or shows that A to (-1, 2) has
required gradient of -2 | 1.1b | A1F
Obtains correct coordinates of D
Or shows that C to (-1,2) has
required gradient of 0.5
Or shows that (-1, 2) lies on
both lines | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 4(b)(i) ---
4(b)(i) | Calculates length of AB and CD.
At least one correct. | 1.1a | M1 | AB = √(36 + 9) = √45 = 3√5
CD = √(4 + 16) = √20 = 2√5
AB + CD = 5√5
Obtains correct simplified sum | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 4(b)(ii) ---
4(b)(ii) | Calculates AD and applies
trapezium area formula | 1.1a | M1 | AD = √(4 + 16) = √20 = 2√5
= (5√5 2√5)
1
2 ×
= 25
Obtains correct area | 1.1b | A1
Subtotal | 2
Question Total | 9
Q | Marking instructions | AO | Marks | Typical solution$ABCD$ is a trapezium where $A$ is the point $(1, -2)$, $B$ is the point $(7, 1)$ and $C$ is the point $(3, 4)$
$DC$ is parallel to $AB$.
$AD$ is perpendicular to $AB$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the equation of the line $CD$.
[2 marks]
\item Show that point $D$ has coordinates $(-1, 2)$
[3 marks]
\end{enumerate}
\item \begin{enumerate}[label=(\roman*)]
\item Find the sum of the length of $AB$ and the length of $CD$ in simplified surd form.
[2 marks]
\item Hence, find the area of the trapezium $ABCD$.
[2 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2021 Q4 [9]}}