| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Intersection of two lines |
| Difficulty | Moderate -0.8 This is a straightforward coordinate geometry question requiring standard techniques: finding the gradient of a perpendicular line, writing its equation, solving simultaneous equations for intersection point P, and calculating a triangle area using the formula ½×base×height. All steps are routine P1 procedures with no conceptual challenges or novel problem-solving required, making it easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3y - 2x = 30 \Rightarrow m = \frac{2}{3}\) | B1 | Gradient of \(l_1 = \frac{2}{3}\) seen or implied |
| \(y = -\frac{3}{2}(x-24)\), or \(y = -\frac{3}{2}x + 36\), or \(2y + 3x = 72\), or \(\frac{y-0}{x-24} = -\frac{3}{2}\) | M1 A1ft | Full method for equation of \(l_2\) using point \((24,0)\) with negative reciprocal gradient. A1ft for correct equation in any form. Follow through on their negative reciprocal gradient |
| Solves simultaneously e.g. \(\frac{2}{3}x + 10 = -\frac{3}{2}(x-24)\) | M1 | Full method to find one coordinate of \(P\) |
| Coordinates of \(P\) \((12, 18)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(l_2\) via \(2y + 3x = c\) | B1 | |
| Full method for equation of \(l_2\) using point \((24,0)\) and \(2y + 3x = c\) | M1 | |
| \(2y + 3x = 72\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 0\), \(3y - 2x = 30 \Rightarrow x = -15\) | M1 | Uses \(y=0\) in \(l_1\) to find \(x\)-coordinate of \(B\). May be implied by \(x = \pm 15\) |
| Area \(BPA = \frac{1}{2} \times (24 + \text{"15"}) \times \text{"18"} = 351\) | dM1 A1 cso | Correct attempt at area using their \(-15\) and \(18\). Depends on first mark and depends on \(B\) being on \(x\)-axis |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3y - 2x = 30 \Rightarrow m = \frac{2}{3}$ | B1 | Gradient of $l_1 = \frac{2}{3}$ seen or implied |
| $y = -\frac{3}{2}(x-24)$, or $y = -\frac{3}{2}x + 36$, or $2y + 3x = 72$, or $\frac{y-0}{x-24} = -\frac{3}{2}$ | M1 A1ft | Full method for equation of $l_2$ using point $(24,0)$ with negative reciprocal gradient. A1ft for correct equation in any form. Follow through on their negative reciprocal gradient |
| Solves simultaneously e.g. $\frac{2}{3}x + 10 = -\frac{3}{2}(x-24)$ | M1 | Full method to find one coordinate of $P$ |
| Coordinates of $P$ $(12, 18)$ | A1 | |
**Alt (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $l_2$ via $2y + 3x = c$ | B1 | |
| Full method for equation of $l_2$ using point $(24,0)$ and $2y + 3x = c$ | M1 | |
| $2y + 3x = 72$ | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 0$, $3y - 2x = 30 \Rightarrow x = -15$ | M1 | Uses $y=0$ in $l_1$ to find $x$-coordinate of $B$. May be implied by $x = \pm 15$ |
| Area $BPA = \frac{1}{2} \times (24 + \text{"15"}) \times \text{"18"} = 351$ | dM1 A1 cso | Correct attempt at area using their $-15$ and $18$. Depends on first mark and depends on $B$ being on $x$-axis |
---5. The line $l _ { 1 }$ has equation $3 y - 2 x = 30$
The line $l _ { 2 }$ passes through the point $A ( 24,0 )$ and is perpendicular to $l _ { 1 }$\\
Lines $l _ { 1 }$ and $l _ { 2 }$ meet at the point $P$
\begin{enumerate}[label=(\alph*)]
\item Find, using algebra and showing your working, the coordinates of $P$.
Given that $l _ { 1 }$ meets the $x$-axis at the point $B$,
\item find the area of triangle $B P A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2021 Q5 [8]}}