How do you write equation of a line goes through (3,1), with slope of m=1/3?

2 Answers
Jan 28, 2017

Use point-slope form #y-y_1=m (x-x_1)#

Explanation:

For this equation that would turn into #y-1=1/3 (x-3)# then solve which should look like #y=1/3x#

Jan 28, 2017

#y=1/3 x#

Explanation:

You know the equation of a line: #y=mx+b#

You are given the point #(3, 1)#. That is the same as saying that #x=3# and #y=1#.

You are given the slope, #m=1/3#. That means that after you "rise" one point on the #y#-axis, you must "run" 3 points on the #x#-axis to get to another point on that line.

Plugging the #x=3#, #y=1#, and #m=1/3# into the equation of a line, you get the following:

#y=mx+b# becomes #1 = 3 * 1/3 + b#. The only variable you don't know yet is #b#. So solve for #b#.

#1 = 3 * 1/3 + b#
#1 = 1 + b#
#b=0#

The final step is to plug just the slope, #m# (previously given), and the #y#-intercept, #b# (which you just found), into the formula #y=mx+b#, giving you:

#y=1/3*x + 0# or simply #y=1/3*x#