The point-slope form of the equation of the line that passes through (–4, –3) and (12, 1) is y – 1 = (x – 12). What is the standard form of the equation for thi

Question

This is hard and I don't understand..please help

Answer 1

Actually, the slope of the line is: 

              m = (y₂-y₁)(x₂-x₁) = (1-(-3))/(12-(-4)) = 4/16 = 1/4

 

So, corrected, the point-slope form of the equation is:

          y-y₁ = m (x-x₁)

          y+3 = (1/4)(x+4)                 [using point (x₁,y₁)]

      or

          y–1 = (1/4)(x–12)                [using point (x₂,y₂]   [let's use this one since you did]

 

To put this into Standard Form (Ax+By=C),

          y–1 = (1/4)(x–12) 

          4y - 4 = x - 12

          -x + 4y - 4 = -12           [subtract x from both sides]

          -x + 4y = -8                 [add 4 to both sides]

           x - 4y = 8            [multiply by (-1) to have x-coefficient positive]

Answer 2

The point-slope form is

 

y-y₁ = m (x-x₁)

 

Your slope is

 

        y₂-y₁      1 - (-3)      4

m = -------- = ---------- = --- = 1/4

        x₂-x₁     12 - (-4)    16

 

So the point-slope equation is

 

y - 1 = (1/4) (x - 12)

 

To get the standard form, you need to solve for y and write the right hand side as mx+b. To do this, we distribute the 1/4 over both terms on the right, then add 1 to both sides:

 

y - 1 = (1/4)x - 3

 

y = (1/4)x - 2

 

So the slope m=1/4 (as before) and the y-intercept, b, is -2.