How do you write the equation of a line in point slope form and slope intercept form given points (6, -4) (-3, 5) ?

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;

#(5-(-4))/(-3-6) = 9/-9 = -1#

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

We should use one point to find the real equation. If we use (-3,5) point;
#y= ax+b => 5=-3a+b#;
a is the slope of the equation, we found that as #-1#;
#5=(-3*-1) + b => 5=3 +b => b=2#;
So the equation of the line will be;
#y=ax+b => ul (y= -x+2)#
We can check if our equation is right or not with other given point;
#(6,-4) => y=-x+2 => -4=-(-1*6)+2 => -4=-6+2 =>-4=-4 #enter image source here
So the equation is correct :)