How do you find a standard form equation for the line with (-4,2) and (6,8)?

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Answer 1 · 2016-08-21T23:26:44

#3x-5y+22=0#

The equation of line joining #(x_1,y_1)# and #(x_2,y_2)# is given by

#(y-y_1)/(y_2-y_1)=(x-x_1)/(x_2-x_1)#.

Hence equation of line joining #(-4,2)# and #(6,8)# is

#(y-2)/(8-2)=(x-(-4))/(6-(-4))# or

#(y-2)/6=(x+4)/10# or

#10y-20=6x+24# or

#6x-10y+24+20=0# or

#6x-10y+44=0# or

#3x-5y+22=0#