How do you write an equation of a line given (-1, 2) and (3, -4)?

2 Answers
Oct 17, 2017

#3x+2y+1=0#

Explanation:

Standard form of equation with two points given is
#(y-y_1)/(y_2-y_1)=(x-x_1)/(x_2-x_1)#
#(y-(-4))/(2-(-4))=(x-3)/(-1-3)#

#(y+4)/6=(x-3)/(-4)#
#-4y-16= 6x - 18#

#6x+4y=2#
#3x+2y=1#

Oct 17, 2017

#y=-3/2x+1/2#

Explanation:

Okay so there are 3 main forms you can use:
- slope-intercept form [#y=mx+b#]
- standard form [#Ax+By=C#]
- point-slope form [#y_1-y_2=m(x_1-x_2)#]

Since you did not specify which form you wanted it in, I am going to use slope-intercept form because that's the easiest to understand, in my opinion. (:

#y=mx+b#

SLOPE (#m#)
To find #m# (slope), you need to find #(rise)/(run)#, or which is the change in #y# divided by the change in #x#. Use this formula: #(y_1-y_2)/(x_1-x_2)#
#(-1, 2) = (x_1, y_1)#
#(3, -4) = (x_2, y_2)#

It doesn't matter which coordinate pair you choose to be #(x_1, y_1)# or #(x_2, y_2)#. Just stay consistent!

#m = [2-(-4)]/[-1-3] = (2+4)/(-1-3) = 6/-4 = -3/2#
#m#= -3/2#

Y-INTERCEPT (#b#)
Choose one of the coordinates to substitute into #y=mx+b#, which will be substituting for #x# and #y#.
I chose #(-1, 2)#. Substitute #m# for #-3/2#.
#(2)=(-3/2)(-1)+b#
Solve for #b#.
#2=3/2+b#
#b=1/2#

FINAL FORM
Substitute #m# and #b# for their values. This is your answer!
#y=-3/2x+1/2#