| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Parameter from distance condition |
| Difficulty | Easy -1.3 This is a straightforward C1 coordinate geometry question testing basic formulas (gradient, midpoint, distance) with minimal problem-solving required. Each part involves direct substitution into standard formulas and simple algebraic manipulation. The most challenging part (iii) requires solving a quadratic equation, but the setup is routine and the numbers work out cleanly. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | uses \(\frac{y_2 - y_1}{x_2 - x_1}\) (at least 3out of 4 correct) | Alternative method: Equation of line through one of the given points with gradient 4 M1 Substitutes other point into their equation M1 Obtains \(p = -1\) (Accept \(y = -1\))A1. Note: Other "informal" methods can score full marks provided www |
| A1 | Correct, unsimplified equation | |
| A1 | ||
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | Correct method (may be implied by one correct coordinate) | Use the same marking principle for candidates who add/subtract half the difference to an end point or use similar triangles or other valid "informal" methods. |
| A1 | ||
| A1 | ||
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| *M1 | Correct method to find line length/square of line length using Pythagoras' theorem (at least 3out of 4 correct) | \((2\sqrt{13})^2 = 52\) or \(2\sqrt{13} = \sqrt{52}\) |
| B1 | ||
| DM1 | Correct method to solve 3 term quadratic, must involve their "52" | SC: B1 for each value of \(d\) found or "spotted" from correct working. Note: Other "informal" methods can score full marks provided www |
| A1 | ||
| [4] |
### (i)
$\frac{p-7}{-4-2}$ or $\frac{7-p}{-2-4}$, $\frac{p-7}{-4-2} = 4$ or $\frac{7-p}{-2-4} = 4$, $p = -1$
| M1 | uses $\frac{y_2 - y_1}{x_2 - x_1}$ (at least 3out of 4 correct) | Alternative method: Equation of line through one of the given points with gradient 4 M1 Substitutes other point into their equation M1 Obtains $p = -1$ (Accept $y = -1$)A1. Note: Other "informal" methods can score full marks provided www |
| A1 | Correct, unsimplified equation | |
| A1 | | |
| [3] | | |
### (ii)
$\frac{-2 + 6}{2} = m$, $\frac{7 + q}{2} = 5$, $m = 2$, $q = 3$
| M1 | Correct method (may be implied by one correct coordinate) | Use the same marking principle for candidates who add/subtract half the difference to an end point or use similar triangles or other valid "informal" methods. |
| A1 | | |
| A1 | | |
| [3] | | |
### (iii)
$\sqrt{(-2-d)^2 + (7-3)^2}$, $d^2 + 4d + 20 = 52$, $d^2 + 4d - 32 = 0$, $(d+8)(d-4) = 0$, $d = -8$ or $4$
| *M1 | Correct method to find line length/square of line length using Pythagoras' theorem (at least 3out of 4 correct) | $(2\sqrt{13})^2 = 52$ or $2\sqrt{13} = \sqrt{52}$ |
| B1 | | |
| DM1 | Correct method to solve 3 term quadratic, must involve their "52" | SC: B1 for each value of $d$ found or "spotted" from correct working. Note: Other "informal" methods can score full marks provided www |
| A1 | | |
| [4] | | |\begin{enumerate}[label=(\roman*)]
\item The line joining the points $(-2, 7)$ and $(-4, p)$ has gradient 4. Find the value of $p$. [3]
\item The line segment joining the points $(-2, 7)$ and $(6, q)$ has mid-point $(m, 5)$. Find $m$ and $q$. [3]
\item The line segment joining the points $(-2, 7)$ and $(d, 3)$ has length $2\sqrt{13}$. Find the two possible values of $d$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 2013 Q6 [10]}}