How do you write the standard form of a line given (12, 1) and has a slope of 4/3?

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Answer 1 · 2016-09-15T10:24:56

#4x-3y-45=0#

The equation of a line given a point #(x_1,y_1)# and slope #m# is given by

#(y-y_1)=m(x-x_1)#

Hence, equation of a line given a point #(12,1)# and slope #4/3# is

#(y-1)=4/3xx(x-12)# or

#3(y-1)=4(x-12)# or

#3y-3=4x-48# or

#4x-3y-45=0#, which is in standard form #ax+by+c=0#