How do you write an equation of a line with points #(5/12, -1), (-3/4, 1/6)#?

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Answer 1 · 2017-03-12T16:01:23

12x+12y+7 = 0

We know equation of a line = #[(y-y_1)/(x-x_1)]=[(y_2-y_1)/(x_2-x_1)]#

here #x_1,y_1 = 5/12, -1 and x_2,y_2= -3/4, 1/6#

hence, #[{y-(-1)}/(x-5/12)] = [{1/6-(-1)}/(-3/4-5/12)]#

#rArr [(y+1)/(x-5/12)] = [(1/6+1)/{(-9-5)/12}]#

#rArr [(y+1)/(x-5/12)] = (7/6)/(-14/12)#

#rArr [(y+1)/(x-5/12)] = 7/6xx-12/14#

#rArr [(y+1)/(x-5/12)] = -1#

#rArr y+1 = -(x-5/12)#

#rArr y+1 +x-5/12 = 0#

#rArr x+y+7/12 = 0#

#rArr 12x+12y+7 = 0#