In order to write the equation of a line in point-slope form, we need the slope of the line, and a point on that line. As we already have a point (in fact, we have two), we need only the slope.
The slope of a line represents the increase (or decrease) in #y# when we increase #x# by #1#. That is, #"slope" = "change in y"/"change in x"#.
To find the slope #m# of a line given two points #(x_1, y_1), (x_2, y_2)# on that line, then, we find the change in #y# as the difference #y_2-y_1#, the change in #x# as the difference #x_2-x_1#, and then divide:
#m = (y_2-y_1)/(x_2-x_1)#
In this case, we have #(x_1,y_1) = (3/2, -1/2)# and #(x_2,y_2)=(-1/2,5/2)#. Plugging that into the above:
#m = (5/2-(-1/2))/(-1/2-3/2)=-3/2#
Then, to get the equation of the line in point-slope form, we can choose one of the points, say, #(3/2, -1/2)#, and plug it and the slope into the format #y - y_1 = m(x-x_1)#:
#y - (-1/2) = -3/2(x-3/2)#
#=> y+1/2 = -3/2(x-3/2)#
Next, to change the equation into standard form, we distribute, then gather the constant term on one side and the variable terms on the other:
#y+1/2 = -3/2x -(-3/2)(3/2)#
#=> y+1/2 = -3/2x + 9/4#
#=> (y + cancel(1/2)) - cancel(1/2) + 3/2x = (cancel(-3/2x) +9/4) - 1/2 + cancel(3/2x)#
#=> y + 3/2x = 9/4 - 1/2#
#:. y+3/2x = 7/4#
