| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Moderate -0.3 This is a standard C1 coordinate geometry question requiring routine techniques: finding gradient from two points, equation of perpendicular line, solving simultaneous equations, distance formula, and triangle area. All steps are straightforward applications of formulas with no novel insight required, making it slightly easier than average but not trivial due to the multi-part nature and final exact answer requirement. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(m = \frac{8-2}{11+1} \left(= \frac{1}{2}\right)\) | M1 A1 | For attempting \(\frac{y_1 - y_2}{x_1 - x_2}\), must be \(y\) over \(x\). No formula condone one sign slip, but if formula is quoted there must be some correct substitution. A1 for fully correct expression, needn't be simplified. |
| \(y - 2 = \frac{1}{2}(x - -1)\) or \(y - 8 = \frac{1}{2}(x - 11)\) o.e. | M1 | For attempting to find equation of \(l_1\) |
| \(y = \frac{1}{2}x + \frac{5}{2}\) | A1 c.a.o. (4) | Accept exact equivalents e.g. \(\frac{6}{12}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Gradient of \(l_2 = -2\) | M1 | For using the perpendicular gradient rule |
| Equation of \(l_2\): \(y - 0 = -2(x - 10)\) \([y = -2x + 20]\) | M1 | For attempting to find equation of \(l_2\). Follow their gradient provided different. |
| \(\frac{1}{2}x + \frac{5}{2} = -2x + 20\) | M1 | For forming a suitable equation to find \(S\) |
| \(x = 7\) and \(y = 6\) | A1, A1 (5) | Depend on all 3 Ms |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(RS^2 = (10-7)^2 + (0-6)^2 (= 3^2 + 6^2)\) | M1 | For expression for \(RS\) or \(RS^2\). Ft their \(S\) coordinates |
| \(RS = \sqrt{45} = 3\sqrt{5}\) \((*)\) | A1 c.s.o. (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(PQ = \sqrt{12^2 + 6^2} = 6\sqrt{5}\) or \(\sqrt{180}\) or \(PS = 4\sqrt{5}\) and \(SQ = 2\sqrt{5}\) | M1, A1 | For expression for \(PQ\) or \(PQ^2\). \(PQ^2 = 12^2 + 6^2\) is M1 but \(PQ = 12^2 + 6^2\) is M0. Allow one numerical slip. |
| \(\text{Area} = \frac{1}{2} PQ \times RS = \frac{1}{2} \cdot 6\sqrt{5} \times 3\sqrt{5}\) | dM1 | For a full, correct attempt at area of triangle. Dependent on previous M1. |
| \(= 45\) | A1 c.a.o. (4) | |
| Total: 15 |
## Question 11:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $m = \frac{8-2}{11+1} \left(= \frac{1}{2}\right)$ | M1 A1 | For attempting $\frac{y_1 - y_2}{x_1 - x_2}$, must be $y$ over $x$. No formula condone one sign slip, but if formula is quoted there must be some correct substitution. A1 for fully correct expression, needn't be simplified. |
| $y - 2 = \frac{1}{2}(x - -1)$ or $y - 8 = \frac{1}{2}(x - 11)$ o.e. | M1 | For attempting to find equation of $l_1$ |
| $y = \frac{1}{2}x + \frac{5}{2}$ | A1 c.a.o. **(4)** | Accept exact equivalents e.g. $\frac{6}{12}$ |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Gradient of $l_2 = -2$ | M1 | For using the perpendicular gradient rule |
| Equation of $l_2$: $y - 0 = -2(x - 10)$ $[y = -2x + 20]$ | M1 | For attempting to find equation of $l_2$. Follow their gradient provided different. |
| $\frac{1}{2}x + \frac{5}{2} = -2x + 20$ | M1 | For forming a suitable equation to find $S$ |
| $x = 7$ and $y = 6$ | A1, A1 **(5)** | Depend on all 3 Ms |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $RS^2 = (10-7)^2 + (0-6)^2 (= 3^2 + 6^2)$ | M1 | For expression for $RS$ or $RS^2$. Ft their $S$ coordinates |
| $RS = \sqrt{45} = 3\sqrt{5}$ $(*)$ | A1 c.s.o. **(2)** | |
### Part (d):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $PQ = \sqrt{12^2 + 6^2} = 6\sqrt{5}$ or $\sqrt{180}$ or $PS = 4\sqrt{5}$ and $SQ = 2\sqrt{5}$ | M1, A1 | For expression for $PQ$ or $PQ^2$. $PQ^2 = 12^2 + 6^2$ is M1 but $PQ = 12^2 + 6^2$ is M0. Allow one numerical slip. |
| $\text{Area} = \frac{1}{2} PQ \times RS = \frac{1}{2} \cdot 6\sqrt{5} \times 3\sqrt{5}$ | dM1 | For a full, correct attempt at area of triangle. Dependent on previous M1. |
| $= 45$ | A1 c.a.o. **(4)** | |
| | **Total: 15** | |\begin{enumerate}
\item The line $l _ { 1 }$ passes through the points $P ( - 1,2 )$ and $Q ( 11,8 )$.\\
(a) Find an equation for $l _ { 1 }$ in the form $y = m x + c$, where $m$ and $c$ are constants.
\end{enumerate}
The line $l _ { 2 }$ passes through the point $R ( 10,0 )$ and is perpendicular to $l _ { 1 }$. The lines $l _ { 1 }$ and $l _ { 2 }$ intersect at the point $S$.\\
(b) Calculate the coordinates of $S$.\\
(c) Show that the length of $R S$ is $3 \sqrt { 5 }$.\\
(d) Hence, or otherwise, find the exact area of triangle $P Q R$.\\
\hfill \mbox{\textit{Edexcel C1 2006 Q11 [15]}}