find the equation of the line which passes through (-8,5) and (4,-1) give your answer in the form ax+by+c=0 where a,b and c are integers?

Question

find the equation of the line which passes through (-8,5) and (4,-1) give your answer in the form ax+by+c=0 where a,b and c are integers?

Answer 1

The approach is to use the point-slope from of the equation of a line, given two points:

    (y-y₁) =          m               (x-x₁)              [where m is the slope of the line]

    (y-y₁) = ( (y₂-y₁)/(x₂-x₁) ) (x-x₁)

  then to convert that into the General Form of a line (ax+by+c=0).

 

    (y-5) = ( (-1-5)/(4-(-8) ) (x-(-8))

     y - 5 = (-6/12)(x+8)

     y - 5 = (-1/2)x - 4

     2y - 10 = -x - 8             [multiply both sides by 2 to get rid of fraction]

     x + 2y -2 = 0               [in ax+by+c=0 form, which is called General Form]

 

Check:

   Is   (-8)+2(5)-2 = 0   ?

         -8+10-2 = 0   ?yes

   Is  4+2(-1)-2 = 0   ?

         4-2-2 = 0   ?

           0 = 0   ?yes

 

 

  

Answer 2

y - (-1) = {[(5 - (-1)] / [(-8) - 4]} ( x - 4 )

 

y + 1 = {[(5+1)/[(-8-4)]}(x-4) ⇔ y = 6/(-12) [x-4] - 1 ⇔ y = -1/2 x + 2 - 1 ⇔ y = -1/2 x + 1 ⇔ 2y + x - 2 = 0 or

 

x + 2y - 2 = 0  as requested.