First of all, the equation of the perpendicular line will have this form:
#y = ax + b#
where #a# is the slope and #b# is the intercept (this is the value of the #y# coordinate when #x=0#).
As we have said, #a# must be inverse to #(-2/5)#, given that both slopes must be inverse. This is a property required because the perpendicularity of both lines: their directions vector are the same, except that their coordinates are exchanged and one of them changes its sign.
#v_"line" = (v_1, v_2) leftrightarrow v_"perpendicular line" = (-v_2, v_1)#
#"slope"_"line" = v_2 / v_1 leftrightarrow "slope"_"perpendicular line" = -v_1 / v_2#
Thus, #a = - (-2/5)^(-1) = - (-5/2) = 5/2#.
Now, let us find the intercept. We know that both lines intercept at #x=1#. So #x=1# must be in the statement line:
#y = - 2/5 cdot 1 - 12 = - 62/5#
So point #(1, -62/5)# is in line #y=-(2x)/5-12#.
Now, this point must be also in perpendicular line, with slope #a = 5/2#.
#y=5/2 x + b rightarrow -62/5 = 5/2 cdot 1 + b rightarrow b = -149/10#
So we have now completed our perpendicular line equation:
#y=5/2 x -149/10#
You can watch the graphics of both lines on this link.
