How do you write the equation of a line in point slope form and slope intercept form given points (-3, 2) and (2, 1)?

1 Answer
May 30, 2015

First we have to find the slope of the equation. To find the slope we have to do;
#(y2-y1)/(x2-x1)# For our question slope is;

#(1-2)/(2-(-3)) = -1/5 #

The main formula of a line is;
#y=ax+b#

We should use one point to find the real equation. If we use (2,1) point;
#y= ax+b => 1=2a+b#;
a is the slope of the equation, we found that as #-1/5#;
#1=(2*-1/5) + b => 1=-2/5 +b => b=1+2/5 =>b=7/5#;
So the equation of the line will be;
#y=ax+b => ul (y= -x/5+7/5)#
We can check if our equation is right or not with other given point;
#(-3,2) => y=-x/5+7/5 => 2=-3*(-1/5)+7/5 => 2=3/5+7/5 => 2=10/5 => 2=2 #enter image source here
So the equation is correct :)