How do you write the equation of the line with a slope of 2/3 and passing through the point (6,-2)?

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Answer 1 · 2016-08-30T11:46:00

We begin by writing #y=2/3x+b#

Now plug in the values for #x# and #y# from the point #(6,-2)#

#-2=2/3*6+b->-2=4+b->b=-6#

So the equation in slope-intercept form is:
#y=2/3x-6#
graph{2/3x-6 [-5.25, 14.75, -7.44, 2.56]}

Answer 2 · 2016-08-30T12:26:13

#y = 2/3x-6#

If we are given the slope and one point, substitute these values into the following formula which is based on the definition for slope.

#y-y_1 = m(x-x_1) color(white)(xxxx) larr m = 2/3 and (6,-2)#

#y -(-2) = 2/3(x -6)color(white)(xxxx)larr# multiply out and simplify.

#y+2 = 2/3x-4#

#y = 2/3x-4-2#

#y = 2/3x-6#