| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Rectangle or parallelogram vertices |
| Difficulty | Standard +0.3 This is a multi-part question requiring standard techniques: finding gradient from line equation, perpendicular gradient, point-slope form, solving simultaneous equations, and using properties of square diagonals. While it has multiple steps (4 parts), each individual technique is routine for P1 level with no novel insight required. The square diagonal property in part (d) adds mild geometric reasoning but remains straightforward application of midpoint formula. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{2}{5}\) or decimal equivalent | B1 | Must be identified; do not accept from rearranged equation unless put in form \(y=\frac{2}{5}x+...\). Do not accept \(\frac{-2}{-5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m_N = -1 \div "\frac{2}{5}"\) | M1 | Correct application of perpendicular gradient rule |
| \(y + 2 = "-\frac{5}{2}"(x-6)\) | M1 | Correct straight-line method with "changed" gradient; allow one sign slip in brackets. If \(y=mx+c\) must proceed to \(c=...\) |
| \(y = -\frac{5}{2}x + 13\) | A1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \("-\frac{5}{2}x+13" = "\frac{2}{5}x+\frac{7}{5}" \Rightarrow "\frac{29}{10}"x = "\frac{58}{5}" \Rightarrow x=...(=4)\) or \("\frac{5}{2}y-\frac{7}{2}" = "-\frac{2}{5}y+\frac{26}{5}" \Rightarrow "\frac{29}{10}"y = "\frac{87}{10}" \Rightarrow y=...(=3)\) | M1 | Correct method to solve for \(x\) or \(y\) from \(l_1\) and \(l_2\). Coordinates found with no algebraic working score 0 |
| \(x="4" \Rightarrow y=...\) or \(y="3" \Rightarrow x=...\) | dM1 | Dependent on previous M |
| \((4, 3)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((2, 8)\) | B1B1 |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2}{5}$ or decimal equivalent | B1 | Must be identified; do not accept from rearranged equation unless put in form $y=\frac{2}{5}x+...$. Do not accept $\frac{-2}{-5}$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m_N = -1 \div "\frac{2}{5}"$ | M1 | Correct application of perpendicular gradient rule |
| $y + 2 = "-\frac{5}{2}"(x-6)$ | M1 | Correct straight-line method with "changed" gradient; allow one sign slip in brackets. If $y=mx+c$ must proceed to $c=...$ |
| $y = -\frac{5}{2}x + 13$ | A1 | oe |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $"-\frac{5}{2}x+13" = "\frac{2}{5}x+\frac{7}{5}" \Rightarrow "\frac{29}{10}"x = "\frac{58}{5}" \Rightarrow x=...(=4)$ or $"\frac{5}{2}y-\frac{7}{2}" = "-\frac{2}{5}y+\frac{26}{5}" \Rightarrow "\frac{29}{10}"y = "\frac{87}{10}" \Rightarrow y=...(=3)$ | M1 | Correct method to solve for $x$ or $y$ from $l_1$ and $l_2$. Coordinates found with no algebraic working score 0 |
| $x="4" \Rightarrow y=...$ or $y="3" \Rightarrow x=...$ | dM1 | Dependent on previous M |
| $(4, 3)$ | A1 | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(2, 8)$ | B1B1 | |8. The line $l _ { 1 }$ has equation
$$2 x - 5 y + 7 = 0$$
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of $l _ { 1 }$
Given that
\begin{itemize}
\item the point $A$ has coordinates $( 6 , - 2 )$
\item the line $l _ { 2 }$ passes through $A$ and is perpendicular to $l _ { 1 }$
\item find the equation of $l _ { 2 }$ giving your answer in the form $y = m x + c$, where $m$ and $c$ are constants to be found.
\end{itemize}
The lines $l _ { 1 }$ and $l _ { 2 }$ intersect at the point $M$.
\item Using algebra and showing all your working, find the coordinates of $M$.\\
(Solutions relying on calculator technology are not acceptable.)
Given that the diagonals of a square $A B C D$ meet at $M$,
\item find the coordinates of the point $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2022 Q8 [9]}}