If you know the slopes of two nonvertical lines,
can you easily determine if they're parallel?
Perpendicular?
YES!
Parallel lines have the same slope.
Perpendicular lines have slopes that are opposite reciprocals.
This section explores these concepts.
Two lines in a plane are parallel if they never intersect;
they have the same ‘slant’.
(For a more rigorous study of parallel lines and related concepts, visit my Geometry course:
Parallel Lines.)
Slope (or lack of it!) can be used to decide if lines are parallel:
This statement offers a good opportunity to review the mathematical words
‘if and only if’
and
‘or’.
The statement has been visually presented to help you see the mathematical sentence structure more clearly.
Now, let's move on to perpendicular lines.
Two lines are perpendicular if they intersect at a $\,90^{\circ}$ angle.
For example, the $\,x$axis and $\,y$axis are perpendicular.
Make sure you understand the opposite reciprocal equation:
$$ \cssId{s61}{m_1 = \frac{1}{m_2}} $$one of the slopes  is  the opposite of  the reciprocal of the other slope 
$\displaystyle \cssId{s63}{m_1}$  $\displaystyle \cssId{s65}{=}$  $\displaystyle \cssId{s67}{}$  $\displaystyle \cssId{s69}{\frac{1}{m_2}}$ 
For example, what's the reciprocal of $\,2\,$?
Answer: $\frac12$
What's the opposite [of the] reciprocal of $\,2\,$?
Answer: $\frac12$
So, the numbers $\,2\,$ and $\,\frac 12\,$ are opposite reciprocals.
Also notice that $\,(2)(\frac 12) = 1\,$.
So, lines with slopes $\,2\,$ and $\,\frac12\,$ are perpendicular.
It's easy to see that this is the correct characterization for perpendicular lines, by studying the sketch below:
The yellow triangle,
with base of length $\,1\,$
and right side of length $\,m\,$,
shows that the slope of the first line is
$\,\cssId{s82}{\frac{\text{rise}}{\text{run}}}
\cssId{s83}{= \frac{m}1}
\cssId{s84}{= m}\,$.
Now, imagine that this yellow triangle is a block of wood that is glued to the line.
Rotate this block of wood counterclockwise by $\,90^{\circ}\,$
(so that the original base is now vertical).
Using the rotated triangle to compute the slope of the new, rotated, line gives:
$\,\cssId{s89}{\frac{\text{rise}}{\text{run}}}
\cssId{s90}{= \frac{1}{m}}
\cssId{s91}{= \frac1m}\,$
Easy! Voila!
CONCEPT QUESTIONS EXERCISE:
On this exercise, you will not key in your answer.However, you can check to see if your answer is correct. 
PROBLEM TYPES:
