How do you write an equation in slope intercept form given that the line passes through the points (4,1) and (2,-3)?

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;

#(-3-1)/(2-4) = (-4)/-2 = 2#

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

We should use one point to find the real equation. If we use (4,1) point;
#y= ax+b => 1=4a+b#;
a is the slope of the equation, we found that as #2#;
#1=(4*2) + b => 1=8 +b => b=-7#;
So the equation of the line will be;
#y=ax+b => ul (y= 2x-7)#
We can check if our equation is right or not with other given point;
#(2,-3) => y=2x-7 => -3=2*2-7 => -3=4-7 => -3=-3 #enter image source here
So the equation is correct :)