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
| Step-by-step | 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={{7}\over{3}}-{{5\cdot x}\over{3}}#. This is equal to #-{{5}\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: \[\begin{array}{rcl}-{{5}\over{3}} \cdot a_l=-1 \\&&\phantom{xxx}\blue{\text{rule perpendicular lines with }a_k=-{{5}\over{3}}} \\ a_l=\frac{-1}{-{{5}\over{3}}} \\&&\phantom{xxx}\blue{\text{both sides divided by }-{{5}\over{3}}} \\ a_l={{3}\over{5}} \\&&\phantom{xxx}\blue{\text{simplified}} \end{array}\] |
| Step 3 | Line #l# is of the form: #y={{3}\over{5}} \cdot x+b#. |
| Step 4 | We fill in point #P# to determine #b#. This gives the equation \[3={{3}\over{5}} \cdot -4+b\] We solve this linear equation for #b# and find \[b={{27}\over{5}}\]. |
| Stap 5 | We fill in the #b# we found in the equation from step #3#. This gives: \[l: y={{3\cdot x}\over{5}}+{{27}\over{5}}\] |
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