### Section 4.3 notes

```Section 4.3
CPM
Your work in section 4.2 did not include writing a linear function in slope
intercept form, but you can certainly do this.
Example: Write the symbolic representation of the linear function with a
slope of  3 and y-intercept of –8.
5
It’s easy, RIGHT!!!!!!! But, what if we are not given the
slope and y-intercept. Perhaps we’re given the slope and
some other point on the line or just two points on the
line. OH NO, what do we do?
We can still use slope-intercept form
y  mx  b
______________________.
The slope is still m, and x
and y can be replaced with the point (or either point if
two are given). Now you just have to solve for the yintercept (b)
Recall: 2 lines are perpendicular iff
they have the opposite reciprical slope
______________________________________.
2 lines are parallel iff
th ey h a ve th e sa m e slo p e
_____________________________.
Example: Write the symbolic representation of a
linear function whose graph has the given
characteristics
a)
passes through the points  2, 4  an d  6, 8 
48
26
m 1
N ow pick a pt to use
 2, 4 
m 1
 2, 4 
y  mx  b
4  1(2)  b
2b
y  1x  2
contains the point (-3, 4) and parallel to –x+3y=150
P arallel lines have sam e slope
– x  3y  150
x
x
so this line has the sam e slope
m 
1
(  3, 4)
3
3 y  x  150
3
y
3
x
3
 50
3
4
1
(  3)  b
3
4  1  b
5b
y
1
3
x5
contains the point (-4,2) and perpendicular to x+4y=16
Perpendicular lines have
the opp recip slope
x  4 y  16
y
x
m 
4
(  4, 2)
1
2  4(  4)  b
2  16  b
4
4
Now flip and switch the
slope
18  b
y  4 x  18
Write an equation of the line that passes through (4,6) and is parallel to
the line that passes through (6,-6) and (10,-4)
So, first we must find the
slope of this line
m 
1
(4, 6)
2
6  4
6  10
6
1
(4)  b
2
2
4
1
2
Now we can use this
slope since they are
parallel
6 2b
4b
y
1
2
x4
```