How do you write an equation of a line that goes through (6,-2); m= -4/3 in standard form?

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Answer 1 · 2015-05-12T22:55:07

Standard form is #y = mx + c# where #m# is the slope and #c# is the #y# coordinate of the intersection of the line with the #y# axis.

We are given #m#, so it remains to determine #c#.

Subtracting #mx# from both sides of the formula yields #c = y - mx#.

Given that the line passes through (6, -2) all we need do is substitute #x = 6# and #y = -2# to get

#c = y - mx#

#= -2 - (-4/3)*6 = -2 + 4*6/3 = -2 + 8 = 6#

So the standard form is #y = -(4/3)x + 6#