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Extended Only Coordinate Geometry

Parallel and Perpendicular Lines for IGCSE Maths

Using gradient relationships for parallel and perpendicular lines. This subtopic is part of Coordinate Geometry in the Cambridge IGCSE Mathematics 0580 syllabus (Extended tier only). Understanding par

What You Need to Know

Using gradient relationships for parallel and perpendicular lines. This subtopic is part of Coordinate Geometry in the Cambridge IGCSE Mathematics 0580 syllabus (Extended tier only). Understanding parallel and perpendicular lines is essential for achieving a strong grade in your IGCSE Maths exam.

Understanding Parallel and Perpendicular Lines

Two lines are parallel if they have the same gradient (m₁ = m₂). Two lines are perpendicular if their gradients multiply to −1 (m₁ × m₂ = −1), meaning each gradient is the negative reciprocal of the other. These facts are tested in IGCSE Extended Paper 4 to find equations of lines, prove geometric properties, and solve coordinate geometry problems. If m₁ = 3/4, the perpendicular gradient is −4/3.

Step-by-Step Method

  1. 1

    Find the gradient of the given line

    Rearrange the given equation into y = mx + c form to read off m. Or calculate m from two given points.

  2. 2

    Determine the required gradient

    Parallel: use the same gradient. Perpendicular: flip and negate the gradient (negative reciprocal). If m = 2, perpendicular gradient = −1/2.

  3. 3

    Find the y-intercept

    Substitute the required gradient and a given point (x₁, y₁) into y = mx + c and solve for c.

  4. 4

    Write the equation

    Substitute m and c into y = mx + c.

  5. 5

    Verify

    For perpendicular lines, check m₁ × m₂ = −1. For parallel lines, check gradients are identical.

Worked Example

Question

Line L has equation y = 3x − 4. Find the equation of the line perpendicular to L that passes through the point (6, 2).

Solution

Step 1: Gradient of L = 3. Step 2: Perpendicular gradient = −1/3 (negative reciprocal of 3). Step 3: Use y = (−1/3)x + c with point (6, 2). 2 = (−1/3)(6) + c 2 = −2 + c c = 4 Step 4: Equation is y = −(1/3)x + 4. Verify: 3 × (−1/3) = −1 ✓ Answer: y = −(1/3)x + 4

Exam Tips for Parallel and Perpendicular Lines

  • Negative reciprocal means flip the fraction AND change the sign: gradient 2/3 becomes −3/2.
  • Check perpendicularity by multiplying the gradients — the product must equal exactly −1.
  • A common error is finding the parallel line instead of perpendicular — reread the question.
  • If a line is horizontal (gradient 0), the perpendicular is vertical (undefined gradient), written as x = constant.

Practice Questions

Q1: Write the equation of the line parallel to y = 2x + 5 that passes through (3, 0).

Show hint

Same gradient (m = 2). Substitute (3, 0): 0 = 2(3) + c → c = −6.

Q2: Show that the lines y = (3/4)x + 1 and y = −(4/3)x − 2 are perpendicular.

Show hint

Multiply the gradients: (3/4) × (−4/3) = −1. This confirms perpendicularity.

Q3: A line passes through A(0, 4) and B(6, 1). Find the equation of the line perpendicular to AB that passes through B.

Show hint

Find gradient of AB = (1−4)/(6−0) = −1/2. Perpendicular gradient = 2. Use point B(6, 1).

Frequently Asked Questions

What is parallel and perpendicular lines in IGCSE Maths?

Using gradient relationships for parallel and perpendicular lines.

Is parallel and perpendicular lines in the Core or Extended syllabus?

Parallel and Perpendicular Lines is part of the Extended only syllabus for IGCSE Mathematics 0580.

How do I revise parallel and perpendicular lines effectively?

Start with the revision notes to understand key concepts, then work through the worked examples step by step. Finally, practise past paper questions under timed conditions. Teacher Rig recommends spending focused revision sessions on parallel and perpendicular lines rather than trying to cover everything at once.

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