Are the lines #y=3x+5# and #y=1/3x+5# perpendicular?

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Answer 1 · 2016-10-09T17:05:52

They are nor perpendicular.

A line can be represented as

#a(x-x_0)+b(y-y_0) = 0# or

#a x+b y -a x_0 - b y_0 = 0# or

calling #p = (x,y), p_0=(x_0,y_0)# and #vec n = (a,b)# we have

#<< p - p_0, vec n >> = 0# for short.

This formulation establishes that in a line passing by #p_0# the locus of #p# is perpendicular to #vec n#.

The lines we have are:

#L_1->y-3x-5=0->(-3)(x-x_0)+(1)(y-y_0) = 0# then
#vec n_1 =(-3,1)# and #p_0=(0,5)#

#L_2->3y-x-15=0->(-1)(x-x_0)+3(y-y_0) = 0# then
#vec n_2 = (-1,3)# and #p_0= (0,5)# now

#<< vec n_1, vec n_2 >> = (-3) xx (-1) + (1) xx (3) = 6# so they are not perpendicular.