| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| 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 straightforward two-part question on perpendicular lines and distance from point to line. Part (a) requires finding the perpendicular gradient and using point-slope form (standard technique), while part (b) involves finding the intersection point and calculating distance using Pythagoras. Both are routine procedures taught early in A-level with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4y + 3x = c\), using \(4 \times 2 + 3 \times 15 = 53\) | M1 (1.1a) | Uses negative reciprocal to obtain equation with correct gradient |
| \(4y + 3x = 53\) e.g. \(y = -\frac{3}{4}x + \frac{53}{4}\) or \(y - 2 = -\frac{3}{4}(x-15)\) | A1 (1.1b) | Obtains correct equation ACF, ISW once ACF seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Solve \(3y - 4x = 21\) and \(4y + 3x = 53\) simultaneously giving \(y = 11\), \(x = 3\) | M1 (1.1a) | Begins elimination of one variable, or obtains correct intersection point |
| \((3-15)^2 + (11-2)^2 = 12^2 + 9^2 = 225\) | M1 (1.1a) | Uses distance formula between \((15, 2)\) and another point, PI correct distance or square of correct distance |
| Distance using \((15, 2)\) and intersection point | M1 (3.1a) | Uses distance formula for \((15, 2)\) and intersection to find distance or \(d^2\) |
| Distance \(= 15\) | A1 (1.1b) | CAO |
## Question 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4y + 3x = c$, using $4 \times 2 + 3 \times 15 = 53$ | M1 (1.1a) | Uses negative reciprocal to obtain equation with correct gradient |
| $4y + 3x = 53$ e.g. $y = -\frac{3}{4}x + \frac{53}{4}$ or $y - 2 = -\frac{3}{4}(x-15)$ | A1 (1.1b) | Obtains correct equation ACF, ISW once ACF seen |
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## Question 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Solve $3y - 4x = 21$ and $4y + 3x = 53$ simultaneously giving $y = 11$, $x = 3$ | M1 (1.1a) | Begins elimination of one variable, or obtains correct intersection point |
| $(3-15)^2 + (11-2)^2 = 12^2 + 9^2 = 225$ | M1 (1.1a) | Uses distance formula between $(15, 2)$ and another point, PI correct distance or square of correct distance |
| Distance using $(15, 2)$ and intersection point | M1 (3.1a) | Uses distance formula for $(15, 2)$ and intersection to find distance or $d^2$ |
| Distance $= 15$ | A1 (1.1b) | CAO |
---5
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the line perpendicular to $L$ which passes through $P$.
5 The line $L$ has equation
5
\item Hence, find the shortest distance from $P$ to $L$.\\
\includegraphics[max width=\textwidth, alt={}, center]{042e248a-9efa-4844-957d-f05715900ffc-05_2488_1716_219_153}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2021 Q5 [6]}}