Download Write Equations of Parallel Lines

2.5
Writing Equation of a Line
Part 2
September 21, 2012
Slopes of Parallel Lines
In a coordinate plane, 2 lines
are parallel if and only if they
have the same slope.
Write Equations of Parallel Lines
Write an equation of the line that passes through( – 1, 4)
and is parallel to the line y = – 2x + 5 .
The given line has a slope of –2. Any line parallel to this
line will also have a slope of –2.
y = mx + b
4 =-2(-1) + b
4 = 2+ b
-2 -2
2=b
y = -2x + 2
Substitute 4 for y, -1 for x and -2 for m
Simplify
Solve for b
Another Example
Find the equation of a line going through the point (3, -5) and parallel to
2
y   x 8
3
Using the point-slope equation where the slope m = -2/3 and
the point is (3, -5) we get
2
y  (5)   ( x  3)
3
2
5   (3)b
2
3
y5   x2
OR
3
 5  2  b
2
y   x 3
3
3  b
2
y   x3
3
Checkpoint
Find the equation of the line going through the point (4,1) and
parallel to y  3x  7 (click mouse for answer)
y  1  3  x  4 
y  1  3 x  12
y  3 x  13
1  3(4)  b
1  12  b
13  b
y  3 x  13
Find the equation of the line going through the point (-2,7) and
parallel to 2 x  y  8
(click mouse for answer)
y  7  2  x    2  
7  2(2)  b
y  7  2  x  2 
y  7  2 x  4
y  2 x  3
7  4b
3b
y  2 x  3
Slopes of Perpendicular Lines
In a coordinate plane, 2 lines are
perpendicular if and only if their
slopes are opposite reciprocal of
each other
(or their product is –1)
Equation of a line Perpendicular to another line
Find the equation of a line going through the point (3, -5) and
perpendicular to
2
y   x8
3
The slope of the perpendicular line will be m = 3/2. Using the point-slope
equation where the slope m = 3/2 and the point (3, -5) we get
3
y  (5)  ( x  3)
2
3
9
y5  x
2
2
3
19
y  x
2
2
3
 5  (3)  b
2
9
5  b
2
19
 b
2
3
19
y  x
2
2
Checkpoint
Find the equation of the line going through the point (-6, -5) and
perpendicular to y = -x + 2.
y = x +1
Find the equation of the line going through the point (-2,7) and
perpendicular to 2 x  y  8
y  7  1 x   2 
2
y7  1
2
x  2
y  7  1 x 1
2
y  1 x8
2

Homework:
2.5 p.98 #37-42ALL, 44-50even