Chapter 1: Linear relations and functions Section 1-5: Writing Equations of

```Chapter 1: Linear relations
and functions
Section 1-5: Writing Equations of
Parallel and Perpendicular lines
Parallel lines have the same slope
Two nonvertical lines in a plane are
parallel if and only if their slopes are
equal and they have no points in
common. Two vertical lines are
always parallel.
m is the same, but the b is different in
the slope-intercept form
Coinciding lines
 Two lines are coinciding if they
represent the same line.
 M and b are the same in the slope
intercept form.
Perpendicular lines have opposite
reciprocal slopes
 Two nonvertical lines in a plane are
perpendicular if and only if their
slopes are opposite reciprocals. A
horizontal and a vertical line are
always perpendicular.
Example # 1 Determine whether the graphs of
each pair of equations are parallel, coinciding,
perpendicular or neither
 2x+4y=4
x+2y=5
Slope-intercept form:
y=- 1 x + 1
2
y=- 1 x + 5
2
Since the slopes are equal
and the b’s are
different, the lines are
parallel
Example #2
 4x+3y-7=0
6x-8y+28=0
Write in slope intercept
form
y= − 4 x + 7
3
3
y= 3 x + 7
4
2
Since the slopes are
opposite reciprocals of
each other, the lines
are perpendicular.
Example #3
 Write the standard form of the equation of the line that
passes through the point at
(-2,10) and is parallel to the graph of 2x+5y+4=0
 Solution: Since we are looking for a line parallel to
2x+5y+4=0, we know that they must have the same slope.
The slope of 2x +5y+4=0 is − 2 .
5

So the equation will have the form y= −
2
x+b .
5
 Now since the point (-2,10) is on this line, it must make the
equation true. In order for that to occur b= 46
2
−
10= 5 (−2) + b
4
so 10= + b
5
5
which means b= 46
Therefore the equation of the line in standard form is
2x+5y-46=0
5
Now you do one.
 Write the standard form of the
equation of the line that passes
through the point at (2,-3), and is
perpendicular to the graph of
6x-8y-5=0
Next Example

Determine the equation of the perpendicular bisector of the line
segment with endpoints A(2,3)and B(6,-1)


Solution
The perpendicular bisector is the line that bisects the existing line at
a right angle
Therefore, we need to find the midpoint of the existing line that has
endpoints (2,3) and (6,-1)
Recall the midpoint formula from geometry.





⎛ 2 + 6 3 + − 1⎞
⎜
⎟
⎜ 2 ,
⎟
2
⎝
⎠
= (4, 1)
Now the slope of the line with points A and B on it is -1, so the slope
of the perpendicular line will be 1 and the equation will be y=x+b.
Now the point (4, 1) is the midpoint and on both lines and will must
make both lines true. S0 1=4+b means that b must be -3. So the
equation of the line is y=x-3. In standard form it is x-y-3=0
HW#5
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Section 1-5
Pp. 36-37
#12,13,15,17,19,21,
23,25,33,37,38,40
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