Straight Lines, Gradients, Midpoint and Length

Q: Can I use y=mx+c for all straight line questions?

Yes, but the point-gradient form is often faster for questions that give you a point on the line and its gradient, as you do not need to solve for the y-intercept c first.

CIE IGCSE Additional Mathematics· Syllabus sections 7.1, 7.2, 7.3· 18 min read

1. 1. Gradient Calculation and Line Classification★★☆☆☆⏱ 4 min

📘 Definition

Gradient

m

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.

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}

📐 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 $m_{1} m_{2} = - 1$ : $(- 1) \times m_{2} = - 1 ⟹ m_{2} = 1$

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. 2. Midpoint and Distance of a Line Segment★★☆☆☆⏱ 5 min

📘 Definition

Midpoint

M

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.

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: $Δ x = - 5 - 3 = - 8$ , $Δ y = 2 - (- 4) = 6$
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. 3. Equation of a Straight Line★★★☆☆⏱ 5 min

📘 Definition

Straight Line Equation

Two standard forms for writing straight line equations: gradient-intercept form ( $y = m x + c$ , 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): $y - 3 = 2 (x - 1) ⟹ y = 2 x + 1$

You can rearrange any straight line equation into the general form $a x + b y + c = 0$ 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 $2 y = 4 x - 7$ . Give your answer in the form $a x + b y + c = 0$ .

1
Step 1: Rearrange given line to find gradient: $2 y = 4 x - 7 ⟹ y = 2 x - 3.5$ , 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: $y - 5 = 2 x - 4 ⟹ 2 x - y + 1 = 0$

Exam tip:

Always check the required form of the answer; losing marks for not rearranging to $a x + b y + c = 0$ is a common avoidable error.

4. 4. Perpendicular Bisector of a Line Segment★★★☆☆⏱ 4 min

📘 Definition

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: $(\frac{1 + 5}{2}, \frac{2 + 6}{2}) = (3, 4)$
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.

5. Common Pitfalls

Wrong move:

Calculating gradient as $\frac{x _{2} - x _{1}}{y _{2} - y _{1}}$ instead of $\frac{y _{2} - y _{1}}{x _{2} - x _{1}}$

Why:

Mixing up rise and run, leading to the reciprocal of the correct gradient

Correct move:

Always remember gradient = rise/run = (change in y)/(change in x)

Wrong move:

Using the parallel gradient rule for perpendicular lines, or vice versa

Why:

Confusing the two rules under exam pressure

Correct move:

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 move:

Forgetting to square the differences when calculating segment length

Why:

Rushing the distance formula, leading to incorrect smaller values for length

Correct move:

Always write the squared terms explicitly in your working before adding them

Wrong move:

Using an endpoint of the original segment instead of the midpoint when calculating the perpendicular bisector

Why:

Misremembering that the bisector must cut the segment exactly in half

Correct move:

First calculate and note the midpoint before finding the perpendicular gradient for the bisector

6. Quick Reference 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	$M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
Segment Length	Find distance between two points	$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
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

7. Frequently Asked

Can I use y=mx+c for all straight line questions?

Yes, but the point-gradient form $y - y_{1} = m (x - x_{1})$ is often faster for questions that give you a point on the line and its gradient, as you do not need to solve for the y-intercept c first.

Do I need to simplify surds for segment length answers?

Unless the question specifies otherwise, yes: leave length answers as simplified surds for non-calculator papers, or rounded to the required number of decimal places for calculator papers.

Going deeper

syllabusCIE 0606 2025-2027 SyllabusOfficial syllabus document for reference

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.