Matched Exercise to Example 1: Find the slope of the line passing through the pairs of points and describe the line as rising, falling, horizontal or vertical.
 (3 , 1) , (3 , 5)
 (1 , 0) , (3 , 7)
 (2 , 1) , (6 , 0)
 (5 , 2) , (9 , 2)
Solution:

The slope of the line is given by
m = ( 5  (1) ) / (3  3)
The denominator is equal to zero and therefore the slope is undefined.

The slope of the line is given by
m = ( 7  0 ) / ( 3  (1) ) = 7 / 4
Since the slope is positive, the line rises as x increases.

We first find the slope of the line
m = ( 0  1 ) / ( 6  2 ) = 1/4
Since the slope is neagtive, the line falls as x increases.

The slope of the line is given by
m = ( 2  2 ) / ( 9  (5) ) = 0
The slope is equal to zero, the line is horizontal.
Matched Exercise to Example 2: A line has a slope
of 5 and passes through the point (1 , 4). Find another point A through which
the line passes. (many possible answers) Solution
Let x_{1} and y_{1} be the x and y coordinates of point A. According to the definition of the slope
( y_{1}  (4) ) / (x_{1}  1) = 5
We need to solve this equation in order to find x_{1} and y_{1}. This equation has two unknowns and therefore has an infinite number of pairs of solutions. We
chose y_{1} and then find x_{1}. if y_{1} = 11, for
example, the above equation becomes
( 11  (4) ) / (x_{1}  1) = 5
We obtain an equation in x_{1}
Solve for x_{1} to obtain
One possible answer is point A at
Check that the two points give a slope of 5
(11  (4) ) / (4  1) = 5
Matched Exercise to Example 3:
Are the lines L1 and L2 passing through the given pairs of points parallel, perpendicular or neither parallel nor perpendicular?
 L1: (1 , 2) , (1 , 1)
L2: (4 , 1) , (4 , 0)
 L1: (2 , 3) , (3 , 1)
L2: (1 , 2) , (7 , 5)
 L1: (1 , 1) , (2 , 2)
L2: (0 , 0) , (1 , 1)
 L1: (1 , 9) , (2 , 9)
L2: (18 , 1) , (0 , 1)
Solution In what follows, m1 is the slope of line L1 and m2
is the slope of line L2.
 m1 = ( 1  2 ) / ( 1  1)
m2 = ( 0  (1) ) / ( 4  (4) )
The two slopes m1 and m2 are undefined since the denominators in m1 and m2 are both equal to zero. Hence the two lines are vertical and therefore parallel
 m1 = ( 1  3 ) / ( 3  2 ) = 2
m2 = ( 5  (2) ) / ( 7  1 ) = 1/2
The product of the two slopes m1*m2 is not equal to 1, the two lines are neither perpendicular nor parallel.
 m1 = ( 2  (1) ) / ( 2  1 ) = 1
m2 = ( 1  0 ) / ( 1  0 ) = 1
The product of m1 and m2 is equal to 1, hence the two line are perpendicular.
 m1 = ( 9  9 ) / ( 2  1 ) = 0
m2 = ( 1  (1) ) / ( 0  18 ) = 0
The two slopes are equal , the two lines
are parallel. Also the two lines are horizontal lines.
Matched Exercise to Example 4: Is it possible for two lines with positive slopes to be perpendicular? Solution
No. If both slopes are positive, their
product can never be equal to 1. 