| Exam Board | Edexcel |
|---|---|
| Module | PURE |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular bisector of segment |
| Difficulty | Moderate -0.8 This is a straightforward coordinate geometry question requiring standard techniques: finding gradient from two points, using y = mx + c form, finding midpoint, using perpendicular gradient rule (negative reciprocal), and converting between line equation forms. All steps are routine A-level procedures with no problem-solving or insight required, making it easier than average but not trivial since it involves multiple connected steps. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Attempts gradient} = \frac{20-(-4)}{-5-(-3)} = (-3)\) | M1 | Attempts gradient = \(\frac{\delta y}{\delta x}\) condoning slips. Look for an attempt at a difference in y coordinates over a difference in x coordinates. There is no requirement to simplify. They may use \(20-(-4) = m(-5-3)\) to find m. |
| \(\text{Attempts equation of line } y-20 = "-3"(x+5) \text{ or } y+4 = "-3"(x-3)\) | dM1 | Full method for equation of line using gradient and either point – as shown or may use \(y = "-3"x + c\) proceeding as far as finding c. |
| \(y = -3x + 5\) | A1 (3) | The coefficients must be simplified, award A0 for \(y = \frac{24}{-8}x+5\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Gradient } \frac{1}{3} \text{ or midpoint } (-1, 8)\) | B1ft | Correct mid point or gradient, allowing follow through on the negative reciprocal of their gradient in (a) |
| \(y - 8 = \frac{1}{3}(x+1)\) | M1 | Attempts the perpendicular bisector (need not be a simplified form). Must be using \(\pm\) the reciprocal of their gradient from (a) with a valid attempt at the midpoint – at least one correct coordinate (or calculator for) must be seen for the midpoint. |
| \(x - 3y + 25 = 0\) | A1 (3) | \(x - 3y + 25 = 0\) but allow any integer multiple and term in a different order (but must be on one side). Ignore if they give a label \(l_2 = x - 3y + 25 = 0\) |
**(a)**
| $\text{Attempts gradient} = \frac{20-(-4)}{-5-(-3)} = (-3)$ | M1 | Attempts gradient = $\frac{\delta y}{\delta x}$ condoning slips. Look for an attempt at a difference in y coordinates over a difference in x coordinates. There is no requirement to simplify. They may use $20-(-4) = m(-5-3)$ to find m. |
|---|---|---|
| $\text{Attempts equation of line } y-20 = "-3"(x+5) \text{ or } y+4 = "-3"(x-3)$ | dM1 | Full method for equation of line using gradient and either point – as shown or may use $y = "-3"x + c$ proceeding as far as finding c. |
| $y = -3x + 5$ | A1 (3) | The coefficients must be simplified, award A0 for $y = \frac{24}{-8}x+5$ |
**(b)**
| $\text{Gradient } \frac{1}{3} \text{ or midpoint } (-1, 8)$ | B1ft | Correct mid point or gradient, allowing follow through on the negative reciprocal of their gradient in (a) |
| $y - 8 = \frac{1}{3}(x+1)$ | M1 | Attempts the perpendicular bisector (need not be a simplified form). Must be using $\pm$ the reciprocal of their gradient from (a) with a valid attempt at the midpoint – at least one correct coordinate (or calculator for) must be seen for the midpoint. |
| $x - 3y + 25 = 0$ | A1 (3) | $x - 3y + 25 = 0$ but allow any integer multiple and term in a different order (but must be on one side). Ignore if they give a label $l_2 = x - 3y + 25 = 0$ |
**Total: 6 marks**
**Guidance for (a):**
- M1: Sets up two simultaneous equations using both points E.g $20 = -5m + c$ and $-4 = 3m + c$ (allow one slip).
- dM1: Solves the pair of simultaneous equations to find values for m and c (accept any values following two suitable equations having been set up).
- A1: $y = -3x + 5$
**Guidance for (b):**
- B1ft: Correct mid point or gradient, allowing follow through on the negative reciprocal of their gradient in (a)
- M1: Attempts the perpendicular bisector (need not be a simplified form). Must be using $\pm$ the reciprocal of their gradient from (a) with a valid attempt at the midpoint – at least one correct coordinate (or calculator for) must be seen for the midpoint.
- A1: $x - 3y + 25 = 0$ but allow any integer multiple and term in a different order (but must be on one side). Ignore if they give a label $l_2 = x - 3y + 25 = 0$
---\begin{enumerate}
\item The line $l _ { 1 }$ passes through the point $A ( - 5,20 )$ and the point $B ( 3 , - 4 )$.\\
(a) Find an equation for $l _ { 1 }$ giving your answer in the form $y = m x + c$, where $m$ and $c$ are constants.
\end{enumerate}
The line $l _ { 2 }$ is perpendicular to $l _ { 1 }$ and passes through the midpoint of $A B$\\
(b) Find an equation for $l _ { 2 }$ giving your answer in the form $p x + q y + r = 0$, where $p , q$ and $r$ are integers.
\hfill \mbox{\textit{Edexcel PURE 2024 Q1}}