Download App
Google Android
Apple iOS
Huawei
English
English
Malay
Guest
Login
Register
Home
Quiz
Battle
Practice
Class
Classes List
Timetable
Assignments
Learn
Learning Hub
Quick Notes
Videos
Experiments
Textbooks
Login
Register
Download App
Google Android
Apple iOS
Huawei
EN
MS
Learn
Quick Notes
List
Solve problems involving equations of parallel and perpendicular lines
Parallel Lines and Perpendicular Lines
7.2
Parallel Lines and Perpendicular Lines
Parallel Lines
Figure
Gradient,
\(m\)
Parallel lines have equal gradients:
\(m_1=m_2\)
.
Example:
The lines
\(y=2x+3\)
and
\(y=2x-4\)
are parallel since they have the same gradient,
\(m=2\)
.
Equation of Parallel Lines
Given a line with equation
\(y=mx+c\)
, any line parallel to it can be written as
\(y=mx+c_1\)
, where
\(c_1\)
is a different constant.
Properties
Parallel lines are always equidistant from each other.
No solution exists for a system of equations representing parallel lines (they do not intersect).
Perpendicular Lines
Figure
Gradient Relationship
If line
\(1\)
has a gradient
\(m_1\)
and line
\(2\)
has a gradient
\(m_2\)
, then
\(m_1 \times m_2=-1\)
.
Example:
The lines
\(y=2x+3\)
and
\(y=-\dfrac{1}{2}x +5\)
are perpendicular because
\(2\times\left(-\dfrac{1}{2}\right)=-1\)
.
Equation of Perpendicular Lines
Given a line with gradient,
\(m\)
, a line perpendicular to it will have a gradient of
\(-\dfrac{1}{m}\)
.
Properties
Perpendicular lines intersect at a
\(90^\circ\)
angle.
At the point of intersection, the gradients of the two lines multiply to
\(-1\)
.
Finding Gradients
From two points
\(m=\dfrac{y_2-y_1}{x_2-x_1}\)
From equations
For a line in the form
\(y=mx+c\)
, the gradient is
\(m\)
.
For a line in the general form
\(ax+by=c\)
, the gradient is
\(m=-\dfrac{a}{b}\)
.
Examples
Parallel Lines
Given line
\(1\)
:
\(y=3x+2\)
, find the equation of a line parallel to it passing through
\((1,4)\)
.
Solution:
SInce the gradient must be the same,
\(m=3\)
.
Using
\(y=mx+c\)
with point
\((1,4)\)
:
\(\begin{aligned} 4&=3(1)+c \\ c&=1. \end{aligned}\)
Equation of parallel line:
\(y=3x+1\)
.
Perpendicular Lines
Given line
\(1\)
:
\(y=2x-5\)
, find the equation of a line perpendicular to it passing through
\((2,3)\)
.
Solution:
Gradient of perpendicular line
\(m_2=-\dfrac{1}{2}\)
.
Using
\(y=mx+c\)
with point
\((2,3)\)
:
\(\begin{aligned} 3&=-\dfrac{1}{2}(2)+c \\ c&=4. \end{aligned}\)
Equation of peprpendicular line:
\(y=-\dfrac{1}{2}x+4\)
.
Parallel Lines and Perpendicular Lines
7.2
Parallel Lines and Perpendicular Lines
Parallel Lines
Figure
Gradient,
\(m\)
Parallel lines have equal gradients:
\(m_1=m_2\)
.
Example:
The lines
\(y=2x+3\)
and
\(y=2x-4\)
are parallel since they have the same gradient,
\(m=2\)
.
Equation of Parallel Lines
Given a line with equation
\(y=mx+c\)
, any line parallel to it can be written as
\(y=mx+c_1\)
, where
\(c_1\)
is a different constant.
Properties
Parallel lines are always equidistant from each other.
No solution exists for a system of equations representing parallel lines (they do not intersect).
Perpendicular Lines
Figure
Gradient Relationship
If line
\(1\)
has a gradient
\(m_1\)
and line
\(2\)
has a gradient
\(m_2\)
, then
\(m_1 \times m_2=-1\)
.
Example:
The lines
\(y=2x+3\)
and
\(y=-\dfrac{1}{2}x +5\)
are perpendicular because
\(2\times\left(-\dfrac{1}{2}\right)=-1\)
.
Equation of Perpendicular Lines
Given a line with gradient,
\(m\)
, a line perpendicular to it will have a gradient of
\(-\dfrac{1}{m}\)
.
Properties
Perpendicular lines intersect at a
\(90^\circ\)
angle.
At the point of intersection, the gradients of the two lines multiply to
\(-1\)
.
Finding Gradients
From two points
\(m=\dfrac{y_2-y_1}{x_2-x_1}\)
From equations
For a line in the form
\(y=mx+c\)
, the gradient is
\(m\)
.
For a line in the general form
\(ax+by=c\)
, the gradient is
\(m=-\dfrac{a}{b}\)
.
Examples
Parallel Lines
Given line
\(1\)
:
\(y=3x+2\)
, find the equation of a line parallel to it passing through
\((1,4)\)
.
Solution:
SInce the gradient must be the same,
\(m=3\)
.
Using
\(y=mx+c\)
with point
\((1,4)\)
:
\(\begin{aligned} 4&=3(1)+c \\ c&=1. \end{aligned}\)
Equation of parallel line:
\(y=3x+1\)
.
Perpendicular Lines
Given line
\(1\)
:
\(y=2x-5\)
, find the equation of a line perpendicular to it passing through
\((2,3)\)
.
Solution:
Gradient of perpendicular line
\(m_2=-\dfrac{1}{2}\)
.
Using
\(y=mx+c\)
with point
\((2,3)\)
:
\(\begin{aligned} 3&=-\dfrac{1}{2}(2)+c \\ c&=4. \end{aligned}\)
Equation of peprpendicular line:
\(y=-\dfrac{1}{2}x+4\)
.
Chapter : Coordinate Geometry
Topic : Solve problems involving equations of parallel and perpendicular lines
Form 4
Additional Mathematics
View all notes for Additional Mathematics Form 4
Related notes
Divisor of a Line Segment
Areas of Polygons
Equations of Loci
Functions
Composite functions
Inverse Functions
Quadratic Equations and Inequalities
Types of Roots of Quadratic Equations
Simultaneous Equations involving One Linear Equation and One Non-Linear Equation
Quadratic Functions
Report this note
Timed Exam
Prepare exams with mock exam papers
Learn more
Register for a free Pandai account now
Edit content
×
Loading...
Quiz
Videos
Notes
Account