| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Moderate -0.8 This is a straightforward C1 coordinate geometry question requiring standard techniques: finding a point on a line, using perpendicular gradient rule (m₁ × m₂ = -1), writing line equations, finding intercepts, and calculating triangle area. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part nature. |
| 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 | Mark | Guidance |
| Gradient of \(l_1 = \frac{4}{5}\) oe | B1 | States or implies gradient of \(l_1 = \frac{4}{5}\). May be implied by perpendicular gradient of \(-\frac{5}{4}\). Do not award for just rearranging to \(y = \frac{4}{5}x + ...\) unless they state \(\frac{dy}{dx} = \frac{4}{5}\) |
| Point \(P = (5, 6)\) | B1 | States or implies \(P\) has coordinates \((5,6)\). \(y=6\) is sufficient. May be seen on diagram. |
| \(-\frac{5}{4} = \frac{y - \text{"6"}}{x-5}\) or \(y - \text{"6"} = -\frac{5}{4}(x-5)\) or \(\text{"6"} = -\frac{5}{4}(5) + c \Rightarrow c = ...\) | M1 | Correct straight line method using \(P(5, \text{"6"})\) and gradient of \(-\frac{1}{\text{grad } l_1}\). Unless \(-\frac{5}{4}\) or \(-\frac{1}{4}\) is being used as gradient here, gradient of \(l_1\) clearly needs to have been identified and its negative reciprocal attempted. |
| \(5x + 4y - 49 = 0\) | A1 | Accept any integer multiple of this equation including "\(= 0\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 0 \Rightarrow 5x + 4(0) - 49 = 0 \Rightarrow x = ...\) or \(y = 0 \Rightarrow 5(0) = 4x + 10 \Rightarrow x = ...\) | M1 | Substitutes \(y=0\) into their \(l_2\) to find a value for \(x\) or substitutes \(y=0\) into \(l_1\) or rearrangement of \(l_1\) to find a value for \(x\). May be implied by a correct value on the diagram. |
| Both \(x\)-intercepts found (Note: at \(T\), \(x = 9.8\) and at \(S\), \(x = -2.5\)) | M1 | Substitutes \(y=0\) into \(l_2\) and \(l_1\) to find both \(x\)-values. May be implied by correct values on diagram. |
| Method 1: \(\frac{1}{2} \times ST \times \text{"6"}\): \(\frac{1}{2} \times (\text{'9.8'} - \text{'-2.5'}) \times \text{'6'} = ...\) | ddM1 | Fully correct method using their values to find area of triangle \(SPT\) with vertices \((5, \text{"6"}), (p,0), (q,0)\) where \(p \neq q\). Attempts to use integration should be sent to team leader. Method 2: \(\frac{1}{2}SP \times PT\): \(\frac{1}{2}\sqrt{(5-\text{'-2.5'})^2+(\text{'6'})^2}\times\sqrt{(\text{'9.8'}-5)^2+(\text{'6'})^2}=...\) Method 3: 2 Triangles: \(\frac{1}{2}\times(5+\text{'2.5'})\times\text{'6'}+\frac{1}{2}\times(\text{'9.8'}-5)\times\text{'6'}=...\) Method 4: Shoelace. Method 5: Trapezium + 2 triangles. Note: if method is correct but slips made when simplifying, method mark can still be awarded. |
| \(= 36.9\) | A1 | 36.9 cso oe e.g. \(\frac{369}{10}\), \(36\frac{9}{10}\), \(\frac{738}{20}\) but not e.g. \(\frac{73.8}{2}\). Final mark is cso so beware of any errors that have fortuitously resulted in a correct area. |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of $l_1 = \frac{4}{5}$ oe | B1 | States or implies gradient of $l_1 = \frac{4}{5}$. May be implied by perpendicular gradient of $-\frac{5}{4}$. Do not award for just rearranging to $y = \frac{4}{5}x + ...$ unless they state $\frac{dy}{dx} = \frac{4}{5}$ |
| Point $P = (5, 6)$ | B1 | States or implies $P$ has coordinates $(5,6)$. $y=6$ is sufficient. **May be seen on diagram.** |
| $-\frac{5}{4} = \frac{y - \text{"6"}}{x-5}$ or $y - \text{"6"} = -\frac{5}{4}(x-5)$ or $\text{"6"} = -\frac{5}{4}(5) + c \Rightarrow c = ...$ | M1 | Correct straight line method using $P(5, \text{"6"})$ and gradient of $-\frac{1}{\text{grad } l_1}$. Unless $-\frac{5}{4}$ or $-\frac{1}{4}$ is being used as gradient here, gradient of $l_1$ clearly needs to have been identified and its negative reciprocal attempted. |
| $5x + 4y - 49 = 0$ | A1 | Accept any integer multiple of this equation including "$= 0$" |
**Total: 4 marks**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 0 \Rightarrow 5x + 4(0) - 49 = 0 \Rightarrow x = ...$ or $y = 0 \Rightarrow 5(0) = 4x + 10 \Rightarrow x = ...$ | M1 | Substitutes $y=0$ into their $l_2$ to find a value for $x$ **or** substitutes $y=0$ into $l_1$ or rearrangement of $l_1$ to find a value for $x$. **May be implied by a correct value on the diagram.** |
| Both $x$-intercepts found (Note: at $T$, $x = 9.8$ and at $S$, $x = -2.5$) | M1 | Substitutes $y=0$ into $l_2$ **and** $l_1$ to find both $x$-values. **May be implied by correct values on diagram.** |
| **Method 1:** $\frac{1}{2} \times ST \times \text{"6"}$: $\frac{1}{2} \times (\text{'9.8'} - \text{'-2.5'}) \times \text{'6'} = ...$ | ddM1 | Fully correct method using their values to find area of triangle $SPT$ with vertices $(5, \text{"6"}), (p,0), (q,0)$ where $p \neq q$. **Attempts to use integration should be sent to team leader.** Method 2: $\frac{1}{2}SP \times PT$: $\frac{1}{2}\sqrt{(5-\text{'-2.5'})^2+(\text{'6'})^2}\times\sqrt{(\text{'9.8'}-5)^2+(\text{'6'})^2}=...$ Method 3: 2 Triangles: $\frac{1}{2}\times(5+\text{'2.5'})\times\text{'6'}+\frac{1}{2}\times(\text{'9.8'}-5)\times\text{'6'}=...$ Method 4: Shoelace. Method 5: Trapezium + 2 triangles. **Note: if method is correct but slips made when simplifying, method mark can still be awarded.** |
| $= 36.9$ | A1 | 36.9 cso oe e.g. $\frac{369}{10}$, $36\frac{9}{10}$, $\frac{738}{20}$ but not e.g. $\frac{73.8}{2}$. **Final mark is cso so beware of any errors that have fortuitously resulted in a correct area.** |
**Total: 4 marks (8 mark question)**
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c1b0a49d-9def-4289-a4cd-288991f67caf-16_659_1438_267_251}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The straight line $l _ { 1 }$, shown in Figure 1, has equation $5 y = 4 x + 10$\\
The point $P$ with $x$ coordinate 5 lies on $l _ { 1 }$\\
The straight line $l _ { 2 }$ is perpendicular to $l _ { 1 }$ and passes through $P$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for $l _ { 2 }$, writing your answer in the form $a x + b y + c = 0$ where $a$, $b$ and $c$ are integers.
The lines $l _ { 1 }$ and $l _ { 2 }$ cut the $x$-axis at the points $S$ and $T$ respectively, as shown in Figure 1.
\item Calculate the area of triangle SPT.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2017 Q8 [8]}}