How do you find the slope given (11, -2) and (2, -2)?
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Slope of a line #m# passing through #(x_1,y_1)# and #(x_2,y_2)# is given by
#" "m=(y_2-y_1)/(x_2-x_1)#
Plugging in the given values we get
#" "m=(11-2)/(-2+2)#
which is undefined, and this gives rise to a special case. The line with undefined slope is a line parallel to the #y#-axis.
Also, slope is #tan theta#. When #tan theta=#undefined, then #theta=pi/2#, which is parallel to #y#-axis.
Nothing, undefined, or #0#
We can find the slope or gradient using:
#(Deltay)/(Deltax)# Where #Delta# means the 'change in...'
Remembering the coordinates are #(x,y)#
#11 -> 2# we have to #-9#
#-2 -> -2 # we do nothing.
Therefore using #(Deltay)/(Deltax)# -> #0/-9 #-> which is nothing, or undefined.
As dividing by #0#, or multiplying by #0# gives nothing.