| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2005 |
| Session | June |
| Marks | 11 |
| 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 multi-part coordinate geometry question requiring routine techniques: rearranging to find gradient, using perpendicular gradient property (m₁m₂ = -1), point-slope form, finding axis intercepts, midpoint formula, and distance formula. While it has four parts worth several marks, each step follows directly from standard procedures with no problem-solving insight required, making it slightly easier than the average A-level question. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \frac{4}{3}x + \frac{5}{3}\), gradient \(= \frac{4}{3}\) | B1 1 | \(\frac{4}{3}\) or 1.33 or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient of \(\perp = -\frac{3}{4}\) | B1ft | \(-\frac{3}{4}\) seen or implied |
| \(y - 2 = -\frac{3}{4}(x-1)\) | M1 | Attempts equation of straight line through \((1,2)\) with any gradient |
| \(4y + 3x = 11\) | A1 | \(y - 2 = -\frac{3}{4}(x-1)\) |
| A1 4 | \(3x + 4y - 11 = 0\) (not aef) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P\left(-\frac{5}{4}, 0\right)\) | B1 | \(\left(-\frac{5}{4}, 0\right)\) seen or implied |
| \(Q\left(0, \frac{11}{4}\right)\) | B1ft | \(\left(0, \frac{11}{4}\right)\) seen or implied (from straight line equation in (ii)) |
| \(\left(-\frac{5}{8}, \frac{11}{8}\right)\) | B1ft 3 | \(\left(-\frac{5}{8}, \frac{11}{8}\right)\) aef |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sqrt{\left(\frac{5}{4}\right)^2 + \left(\frac{11}{4}\right)^2}\) | M1 | Correct method to find line length using Pythagoras' theorem |
| \(\frac{\sqrt{146}}{4}\) | A1 | \(\sqrt{\left(\frac{5}{4}\right)^2 + \left(\frac{11}{4}\right)^2}\) |
| A1 3 | \(\frac{\sqrt{146}}{4}\) |
## Question 9(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \frac{4}{3}x + \frac{5}{3}$, gradient $= \frac{4}{3}$ | B1 **1** | $\frac{4}{3}$ or 1.33 or better |
## Question 9(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of $\perp = -\frac{3}{4}$ | B1ft | $-\frac{3}{4}$ seen or implied |
| $y - 2 = -\frac{3}{4}(x-1)$ | M1 | Attempts equation of straight line through $(1,2)$ with any gradient |
| $4y + 3x = 11$ | A1 | $y - 2 = -\frac{3}{4}(x-1)$ |
| | A1 **4** | $3x + 4y - 11 = 0$ (not aef) |
## Question 9(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P\left(-\frac{5}{4}, 0\right)$ | B1 | $\left(-\frac{5}{4}, 0\right)$ seen or implied |
| $Q\left(0, \frac{11}{4}\right)$ | B1ft | $\left(0, \frac{11}{4}\right)$ seen or implied (from straight line equation in (ii)) |
| $\left(-\frac{5}{8}, \frac{11}{8}\right)$ | B1ft **3** | $\left(-\frac{5}{8}, \frac{11}{8}\right)$ aef |
## Question 9(iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sqrt{\left(\frac{5}{4}\right)^2 + \left(\frac{11}{4}\right)^2}$ | M1 | Correct method to find line length using Pythagoras' theorem |
| $\frac{\sqrt{146}}{4}$ | A1 | $\sqrt{\left(\frac{5}{4}\right)^2 + \left(\frac{11}{4}\right)^2}$ |
| | A1 **3** | $\frac{\sqrt{146}}{4}$ |
---9 (i) Find the gradient of the line $l _ { 1 }$ which has equation $4 x - 3 y + 5 = 0$.\\
(ii) Find an equation of the line $l _ { 2 }$, which passes through the point ( 1,2 ) and which is perpendicular to the line $l _ { 1 }$, giving your answer in the form $a x + b y + c = 0$.
The line $l _ { 1 }$ crosses the $x$-axis at $P$ and the line $l _ { 2 }$ crosses the $y$-axis at $Q$.\\
(iii) Find the coordinates of the mid-point of $P Q$.\\
(iv) Calculate the length of $P Q$, giving your answer in the form $\frac { \sqrt { } a } { b }$, where $a$ and $b$ are integers.
\hfill \mbox{\textit{OCR C1 2005 Q9 [11]}}