How do you find the point-slope form of the equation of the line passing through the points (8,11), (6,16)?

1 Answer
Elvan Burcu K.
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;

#(16-11)/(6-8) = 5/-2 = -5/2#

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

We should use one point to find the real equation. If we use (6,16) point;
#y= ax+b => 16=6a+b#;
a is the slope of the equation, we found that as #-5/2#;
#16=(6*-5/2) + b => 16=-15 +b => b=31#;
So the equation of the line will be;
#y=ax+b => ul (y= -5/2x+31)#
We can check if our equation is right or not with other given point;
#(8,11) => y=-5/2+31 => 11=(-5/2*8)+31 => 11=-20+31 => 11=11 #enter image source here
So the equation is correct :)

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