How do you write an equation of a line through (-3,-1) and (1/2,2)?

1 Answer
Jun 13, 2017

The equation of that line is #y =(7x + 15)/6#.

Explanation:

The equation of a line is based on two simple questions: "How much #y# changes when you add #1# to #x#?" and "How much is #y# when #x=0#?"

First, it's important to know that a linear equation has a general formula defined by #y = m*x + n#.

Having those questions in mind, we can find the slope (#m#) of the line, that is how much #y# changes when you add #1# to #x#:
#m = (D_x)/(D_y)#, with #D_x# being the difference in #x# and #D_y# being the difference in #y#.

#D_x = -3-(1/2) = -3-1/2 = -7/2#
#D_y = -1-(2) = -1-2 = -3#

#m = (-7/2)/-3 = -7/2 * -1/3 = (-7*-1)/(2*3) = 7/6#

Now, we need to find #y_0#, that is the value of #y# when #x=0#:
Since when #x=-3#, #y=-1#, we can sum the slope #3# times (since #-3 +3 = 0#) in #y#:

#y_0 = -1 + 3*(7/6) = -1+21/6 = - 6/6+21/6 = 15/6#

We now have the slope and the #y_0# (or #n#) value, we apply in the main formula of a linear equation:

#y = m*x + n = 7/6 * x + 15/6 = (7x + 15)/6#