What is the equation of the line passing through #(88,93)# and #(-120,3)#?

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Answer 1 · 2018-03-03T11:34:06

The equation of the line is #45x-104y= -5712#

The slope of the line passing through #(88,93) and (-120,3)#

is #m= (y_2-y_1)/(x_2-x_1)= (3-93)/(-120-88)= 90/208=45/104#

Let the equation of the line in slope-intercept form be #y=mx+c#

#:. y=45/104x+c#. The point #(88,93)# will satisfy the equation .

,# :. 93= 45/104*88+c or 104*93=45*88+104c# or

#104c=104*93-45*88or c=(104*93-45*88)/104# or

#c= 5712/104=1428/26=714/13#

Hence the equation of the line is #y= 45/104x+714/13# or

#104y = 45x+5712 or 45x-104y= -5712# [Ans]