How do you find a standard form equation for the line with slope 2/3 that passes through the point (3,6)?

The statement gives that #m=2/3#, so we have to find the value of #b#

Now, we also know that the line passes through #(3, 6)#, and so:

#6 = 2/3*3+ b# and then #6=2 + b# and so #b=4#. The equation of the line is then:

#y=2/3 x + 4#

Start by finding the slope-intercept form of the equation, then converting that to the standard form.

The slope-intercept form is
#y = mx + b#

The slope #m# is given as #2/3#, so the equation up to that point is
#y = (2)/(3)x + b#

To find #b#, sub in the values for #x# and #y# from the ordered pair given in the problem.

#6 = (2)/(3)((3)/(1)) + b#
Solve for #b#

1) Clear the parentheses by multiplying the fractions
#6 = 2 + b#

2) Subtract #2# from both sides to isolate #b#
#4 = b#

So the slope-intercept form of the equation is
#y = (2/3)x + 4# #larr# slope-intercept form

Change the slope-intercept form into standard form.

Standard form is
#ax + by = c# where #a# is a positive whole digit

1) Clear the fraction by multiplying all the terms on both sides by #3# and letting the denominator cancel
#3y = 2x + 12#

2) Subtract #2x# from both sides to get the #x# and #y# terms on the same side
#- 2x + 3y = 12#

3) Multiply through by #-1# to clear the minus sign
#2x - 3y = - 12# #larr# standard form