CIE IGCSE Mathematics (0580) Perpendicular lines Study Notes - New Syllabus
CIE IGCSE Mathematics (0580) Perpendicular lines Study Notes
LEARNING OBJECTIVE
- Perpendicular Lines
Key Concepts:
- Perpendicular Lines
Perpendicular Lines
Perpendicular Lines
Two lines are said to be perpendicular if they meet at a right angle (90°). In coordinate geometry, perpendicular lines have slopes (gradients) that are negative reciprocals of each other.
Key Concept:
- If one line has a gradient \( m \), then a line perpendicular to it will have a gradient of \( -\frac{1}{m} \).
- This only applies when \( m \neq 0 \) and the line is not vertical (since vertical and horizontal lines are also perpendicular).
Steps to Find Equation of Perpendicular Line:
- Find the gradient of the original line.
- Use the negative reciprocal to get the new gradient.
- Use the point the new line passes through and apply the formula \( y – y_1 = m(x – x_1) \).
Example :
Find the equation of the line perpendicular to \( y = 3x + 2 \) and passing through the point (4, 1).
▶️ Answer/Explanation
Original gradient = 3
Perpendicular gradient = \( -\frac{1}{3} \)
Using point-slope form: \( y – y_1 = m(x – x_1) \)
\( y – 1 = -\frac{1}{3}(x – 4) \)
\( y = -\frac{1}{3}x + \frac{4}{3} + 1 = -\frac{1}{3}x + \frac{7}{3} \)
Answer: \( y = -\frac{1}{3}x + \frac{7}{3} \)
Example :
A line has the equation \( 2x + y = 5 \). Find the equation of the line perpendicular to it that passes through (0, 3).
▶️ Answer/Explanation
First rearrange the given equation into slope-intercept form:
\( y = -2x + 5 \), so gradient = -2
Perpendicular gradient = \( \frac{1}{2} \)
Using point-slope form with (0, 3):
\( y – 3 = \frac{1}{2}(x – 0) \Rightarrow y = \frac{1}{2}x + 3 \)
Answer: \( y = \frac{1}{2}x + 3 \)
Example:
Find the equation of the perpendicular bisector of the line joining the points (–3, 8) and (9, –2).
▶️ Answer/Explanation
Midpoint = \( \left( \frac{-3 + 9}{2}, \frac{8 + (-2)}{2} \right) = \left( \frac{6}{2}, \frac{6}{2} \right) = (3, 3) \)
Gradient = \( \frac{-2 – 8}{9 – (-3)} = \frac{-10}{12} = -\frac{5}{6} \)
Perpendicular gradient = \( -\frac{1}{\left(-\frac{5}{6}\right)} = \frac{6}{5} \)
Using point (3, 3) and gradient \( \frac{6}{5} \):
\( y – 3 = \frac{6}{5}(x – 3) \)
Multiply through: \( 5(y – 3) = 6(x – 3) \)
\( 5y – 15 = 6x – 18 \)
Final equation: \( 6x – 5y = 3 \)
Answer: \( 6x – 5y = 3 \)