Geometry: Lines
Perpendicular lines
We have already seen that two perpendicular lines make an angle of #90^\circ# or #\frac{\pi}{2}# radians. This gives a relation between the slopes of two perpendicular lines.
Perpendicular Lines
For two lines #\blue k# and #\green l# with slope #a_{\blue k}# and #a_{\green l}# we have:
\[\begin{array}{c} a_{\blue k} \cdot a_{\green l}=1 \\ \text{ if and only if }\\ \text{line }\blue k \text{ and line } \green l \text{ are perpendicular}\end{array}\]
This means that whenever #a_{\blue k} \cdot a_{\green l}=1# the lines are perpendicular.
And also that if the lines #\blue k# and #\green l# are perpendicular then #a_{\blue k} \cdot a_{\green l}=1#.
Given a line #k# and a point #P# we can use this to determine a line #l# perpendicular to #k# that passes through #P#.
Determining Perpendicular Line
Stepbystep  Example  
We determine the line #\green l# perpendicular to a line #\blue k# that passes through a point #P#. 
#\blue k: y=3x+5# #P=\rv{3,2}# 

Step 1 
Determine the slope of line #\blue k#. 
#a_{\blue k}=3# 
Step 2 
Determine the slope of line #\green l# using the rule #a_{\blue k} \cdot a_{\green l}=1#. 
#a_{\green l}=\frac{1}{3}# 
Step 3 
The equation of line #\green l# is of the form \[y=a_{\green l} \cdot x+b\] 
#\green l: y=\frac{1}{3}x+b# 
step 4 
Determine #b# by substituting the coordinates of point #P# and solving the resulting equation. 
#b=3# 
step 5 
Substitute #b# in the equation from step 3. 
#\green l: y=\frac{1}{3}x+3# 
Step 1  We determine the slope of line #k: y={{1}\over{9}}{{2\cdot x}\over{3}}#. This is equal to #{{2}\over{3}}#. 
Step 2  We now determine the slope of line #l# with the rule: #a_k \cdot a_l=1#. This goes as follows: \[\begin{array}{rcl}{{2}\over{3}} \cdot a_l=1 \\&&\phantom{xxx}\blue{\text{rule perpendicular lines with }a_k={{2}\over{3}}} \\ a_l=\frac{1}{{{2}\over{3}}} \\&&\phantom{xxx}\blue{\text{both sides divided by }{{2}\over{3}}} \\ a_l={{3}\over{2}} \\&&\phantom{xxx}\blue{\text{simplified}} \end{array}\] 
Step 3  Line #l# is of the form: #y={{3}\over{2}} \cdot x+b#. 
Step 4  We fill in point #P# to determine #b#. This gives the equation \[2={{3}\over{2}} \cdot 1+b\] We solve this linear equation for #b# and find \[b={{7}\over{2}}\]. 
Stap 5  We fill in the #b# we found in the equation from step #3#. This gives: \[l: y={{3\cdot x}\over{2}}{{7}\over{2}}\] 
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