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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
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