What is the equation of the line that passes through #(2,- 1)# and #( - 10,4)#?

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Answer 1 · 2017-05-14T17:54:50

#y-(-1)=-5/12(x-2)# or #y=-5/12x-2/12#

First, find the slope:

Slope is defined as #m=(y_2-y_1)/(x_2-x_1)#

It doesn't really matter which you call #(x_1,y_1)#. I'll just call the first one that. So:

#m=(4-(-1))/(-10-2)=5/-12#

So now we have the slope. We can plug into the point-slope form which is:

#y-y_1=m(x-x_1)#

Again, it doesn't really matter what you call #(x_1,y_1)#. I'll call the first one that:

#y-(-1)=-5/12(x-2)#

You could leave it like this but I assume you would like it in slope intercept form which is #y=mx+b#. To do this, solve for #y#

#y+1=-5/12x+10/12#

#y=-5/12x-2/12#

#y=-5/12x-1/6#