| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Distance between two points |
| Difficulty | Easy -1.2 This is a straightforward multi-part question testing basic coordinate geometry: finding a line equation from point and gradient (standard y-y₁=m(x-x₁)), verifying a point lies on a line (substitution), distance formula application, and setting up a distance equation. All parts are routine textbook exercises requiring only direct application of standard formulas with no problem-solving insight needed. The algebraic manipulation in part (d) is minimal. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y - 5 = -\frac{1}{2}(x-2)\) or equivalent e.g. \(\frac{y-5}{x-2} = -\frac{1}{2}\) | M1, A1 | M1A0 if 5 and 2 are wrong way round but correct formula seen |
| \(y = -\frac{1}{2}x + 6\) | A1cao | Correct answer without working scores full marks (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = -2 \Rightarrow y = -\frac{1}{2}(-2) + 6 = 7\) (therefore \(B\) lies on line) | B1 | Conclusion not required except when method used is to establish line through \((-2,7)\) with gradient \(-\frac{1}{2}\) has same equation (1 mark total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(AB^2 = (2-(-2))^2 + (7-5)^2 = 16 + 4 = 20\) | M1, A1 | M1 for attempting \(AB^2\) or \(AB\); allow one slip inside bracket; do not allow \((2-(-2))^2 - (7-5)^2\) |
| \(AB = \sqrt{20} = 2\sqrt{5}\) | A1 | 1st A1 for 20; 2nd A1 for \(2\sqrt{5}\) or \(k=2\) (ignore \(\pm\)) (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(C\) is \(\left(p,\ -\frac{1}{2}p + 6\right)\), so \(AC^2 = (p-2)^2 + \left(-\frac{1}{2}p + 6 - 5\right)^2\) | M1 | For \((p-2)^2 + (\text{linear function of } p)^2\) |
| \(25 = p^2 - 4p + 4 + \frac{1}{4}p^2 - p + 1\) | M1 | Dependent on 1st M1; for forming equation in \(p\) using 25 or 5 and attempting to multiply out both brackets |
| \(25 = 1.25p^2 - 5p + 5\) or \(100 = 5p^2 - 20p + 20\) | A1 | For collecting like \(p\) terms and having correct expression |
| \(0 = p^2 - 4p - 16\) | A1cso | For correct work leading to printed answer (4 marks total) |
# Question 10:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y - 5 = -\frac{1}{2}(x-2)$ or equivalent e.g. $\frac{y-5}{x-2} = -\frac{1}{2}$ | M1, A1 | M1A0 if 5 and 2 are wrong way round but correct formula seen |
| $y = -\frac{1}{2}x + 6$ | A1cao | Correct answer without working scores full marks (3 marks total) |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = -2 \Rightarrow y = -\frac{1}{2}(-2) + 6 = 7$ (therefore $B$ lies on line) | B1 | Conclusion not required except when method used is to establish line through $(-2,7)$ with gradient $-\frac{1}{2}$ has same equation (1 mark total) |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $AB^2 = (2-(-2))^2 + (7-5)^2 = 16 + 4 = 20$ | M1, A1 | M1 for attempting $AB^2$ or $AB$; allow one slip inside bracket; do not allow $(2-(-2))^2 - (7-5)^2$ |
| $AB = \sqrt{20} = 2\sqrt{5}$ | A1 | 1st A1 for 20; 2nd A1 for $2\sqrt{5}$ or $k=2$ (ignore $\pm$) (3 marks total) |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $C$ is $\left(p,\ -\frac{1}{2}p + 6\right)$, so $AC^2 = (p-2)^2 + \left(-\frac{1}{2}p + 6 - 5\right)^2$ | M1 | For $(p-2)^2 + (\text{linear function of } p)^2$ |
| $25 = p^2 - 4p + 4 + \frac{1}{4}p^2 - p + 1$ | M1 | Dependent on 1st M1; for forming equation in $p$ using 25 or 5 and attempting to multiply out both brackets |
| $25 = 1.25p^2 - 5p + 5$ or $100 = 5p^2 - 20p + 20$ | A1 | For collecting like $p$ terms and having correct expression |
| $0 = p^2 - 4p - 16$ | A1cso | For correct work leading to printed answer (4 marks total) |\begin{enumerate}
\item The line $l _ { 1 }$ passes through the point $A ( 2,5 )$ and has gradient $- \frac { 1 } { 2 }$.\\
(a) Find an equation of $l _ { 1 }$, giving your answer in the form $y = m x + c$.
\end{enumerate}
The point $B$ has coordinates (-2, 7).\\
(b) Show that $B$ lies on $l _ { 1 }$.\\
(c) Find the length of $A B$, giving your answer in the form $k \sqrt { } 5$, where $k$ is an integer.
The point $C$ lies on $l _ { 1 }$ and has $x$-coordinate equal to $p$.\\
The length of $A C$ is 5 units.\\
(d) Show that $p$ satisfies
$$p ^ { 2 } - 4 p - 16 = 0 .$$
\hfill \mbox{\textit{Edexcel C1 2009 Q10 [11]}}