# Straight Lines, Gradients, Midpoint and Length

> CIE IGCSE Additional Mathematics · 0606 (2025-2027)
> Source: https://www.owlsprep.com/study/cie-0606-u7-straight-lines-gradients-midpoint-and/

This guide covers all core straight line coordinate geometry skills required for CIE IGCSE Additional Mathematics 0606, including gradient calculation, segment length, midpoint, parallel/perpendicular rules and perpendicular bisectors.

**Prerequisites:** Basic 2D Cartesian coordinate plotting; Algebraic rearrangement of linear equations

## Learning objectives

- Calculate the gradient of a straight line between two coordinate points
- Apply midpoint and distance formulas to solve line segment problems
- Identify parallel and perpendicular lines using gradient rules
- Derive straight line equations in gradient-intercept, point-gradient and general form
- Find the perpendicular bisector of a given line segment

## 1. Gradient Calculation and Line Classification

**Gradient** — Measures the steepness of a straight line, calculated as the change in y divided by the change in x between two points on the line.

*Notation:* m

*Example:* Gradient between (1,2) and (3,6) is $\frac{6-2}{3-1} = 2$

The gradient formula for two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

> **info**
>
> Two lines are parallel if and only if their gradients are equal ($m_1 = m_2$). Two lines are perpendicular if the product of their gradients is $-1$ ($m_1 m_2 = -1$), excluding vertical/horizontal pairs where gradients are undefined and 0 respectively.

**Worked example:** Find the gradient of the line passing through A(-2, 5) and B(4, -1), and state the gradient of a line perpendicular to AB.

1. Step 1: Label coordinates: $x_1 = -2, y_1 =5, x_2=4, y_2=-1$
2. $$m = \frac{-1 - 5}{4 - (-2)} = \frac{-6}{6} = -1$$
3. Step 3: For perpendicular gradient, use $m_1 m_2 = -1$: $(-1) \times m_2 = -1 \implies m_2 = 1$

> **Exam tip:** Always double-check subtraction order when calculating gradient: reversing both x and y differences gives the same result, but reversing only one gives the negative of the correct value.

## 2. Midpoint and Distance of a Line Segment

**Midpoint** — Point exactly halfway between two endpoints of a line segment, with coordinates equal to the average of the x-coordinates and average of the y-coordinates of the endpoints.

*Notation:* M

*Example:* Midpoint of (2,3) and (6,7) is (4,5)

$$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$

The length of a line segment between two points is found using Pythagoras' theorem, since the difference in x and difference in y form the two shorter sides of a right triangle with the segment as hypotenuse:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

**Worked example:** Calculate the midpoint and length of the line segment between P(3, -4) and Q(-5, 2). Give length to 2 decimal places.

1. Step 1: Calculate midpoint: average x = $\frac{3 + (-5)}{2} = -1$, average y = $\frac{-4 + 2}{2} = -1$, so midpoint = $(-1, -1)$
2. Step 2: Calculate differences: $\Delta x = -5 -3 = -8$, $\Delta y = 2 - (-4) = 6$
3. $$d = \sqrt{(-8)^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10.00$$

> **Exam tip:** For non-calculator papers, squared terms in the distance formula will always produce perfect squares or simple surds; if you get an ugly number, you have likely made an arithmetic error.

## 3. Equation of a Straight Line

**Straight Line Equation** — Two standard forms for writing straight line equations: gradient-intercept form ($y=mx+c$, where c is the y-intercept) and point-gradient form ($y-y_1 = m(x-x_1)$, used when you know a point on the line and its gradient).

*Example:* Line with gradient 2 passing through (1,3): $y - 3 = 2(x -1) \implies y = 2x +1$

You can rearrange any straight line equation into the general form $ax + by + c =0$ where $a, b, c$ are integers, if required by the question.

**Worked example:** Find the equation of the line passing through (2, 5) that is parallel to the line $2y = 4x -7$. Give your answer in the form $ax + by + c =0$.

1. Step 1: Rearrange given line to find gradient: $2y = 4x -7 \implies y = 2x - 3.5$, so $m=2$. Parallel lines have equal gradient, so required line gradient is 2.
2. Step 2: Use point-gradient form with $(x_1,y_1)=(2,5)$ and $m=2$: $y -5 = 2(x -2)$
3. Step 3: Rearrange to general form: $y -5 = 2x -4 \implies 2x - y +1 = 0$

> **Exam tip:** Always check the required form of the answer; losing marks for not rearranging to $ax + by + c = 0$ is a common avoidable error.

## 4. Perpendicular Bisector of a Line Segment

**Perpendicular Bisector** — A line that passes through the midpoint of a line segment and is perpendicular to it, meaning all points on the bisector are equidistant from the two endpoints of the segment.

To find the perpendicular bisector of a segment between two points, you need three pieces of information: the midpoint of the segment, the gradient of the original segment, and the perpendicular gradient to use for the bisector.

**Worked example:** Find the perpendicular bisector of the line segment joining A(1, 2) and B(5, 6). Give your answer in gradient-intercept form.

1. Step 1: Find midpoint of AB: $\left(\frac{1+5}{2}, \frac{2+6}{2}\right) = (3,4)$
2. Step 2: Find gradient of AB: $\frac{6-2}{5-1} = 1$, so perpendicular gradient = $-1$
3. Step 3: Use point-gradient form with midpoint (3,4) and $m=-1$: $y -4 = -1(x -3)$
4. Step 4: Rearrange to gradient-intercept form: $y = -x + 7$

> **Exam tip:** The perpendicular bisector always passes through the midpoint of the original segment, so use this to check your answer by substituting the midpoint into your final equation.

## Common pitfalls

- **Wrong:** Calculating gradient as $\frac{x_2 - x_1}{y_2 - y_1}$ instead of $\frac{y_2 - y_1}{x_2 - x_1}$
  - Why it fails: Mixing up rise and run, leading to the reciprocal of the correct gradient
  - Correct: Always remember gradient = rise/run = (change in y)/(change in x)
- **Wrong:** Using the parallel gradient rule for perpendicular lines, or vice versa
  - Why it fails: Confusing the two rules under exam pressure
  - Correct: Write down $m_1 = m_2$ for parallel and $m_1 m_2 = -1$ for perpendicular at the start of any line classification question
- **Wrong:** Forgetting to square the differences when calculating segment length
  - Why it fails: Rushing the distance formula, leading to incorrect smaller values for length
  - Correct: Always write the squared terms explicitly in your working before adding them
- **Wrong:** Using an endpoint of the original segment instead of the midpoint when calculating the perpendicular bisector
  - Why it fails: Misremembering that the bisector must cut the segment exactly in half
  - Correct: First calculate and note the midpoint before finding the perpendicular gradient for the bisector

## Cheatsheet

| Formula/Rule | Use Case | Notation |
| --- | --- | --- |
| Gradient | Find steepness between two points | $m = \frac{y_2 - y_1}{x_2 - x_1}$ |
| Parallel Lines | Check if lines are parallel | $m_1 = m_2$ |
| Perpendicular Lines | Check if lines are perpendicular | $m_1 m_2 = -1$ |
| Midpoint | Find halfway point of a segment | $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ |
| Segment Length | Find distance between two points | $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ |
| Point-Gradient Equation | Write line equation from point + gradient | $y - y_1 = m(x - x_1)$ |
| Perpendicular Bisector | Line cutting segment in half at 90° | Uses midpoint + perpendicular gradient |

## What's next

Now that you have mastered core straight line coordinate geometry skills, you can apply these concepts to more advanced problems involving intersections of lines, graphical solution of simultaneous equations, and connecting coordinate geometry to vector and circle questions later in the CIE IGCSE Additional Mathematics 0606 syllabus. Practice applying these formulas to mixed questions to build speed and accuracy for both calculator and non-calculator papers, as these skills form the foundation for many higher-mark coordinate geometry problems.

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