Question

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 this line?

Answer 1

      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

`x-4y=8`

Explanation:

Given : (–4, –3) , (12, 1)

Solution:

`(x_1,y_1)=(12,1)`

`(x_2,y_2)=(-4,-3)`

Two point slope form :
`y-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)`

Substitute the given values.

`y-1=(-3-1)/(-4-12)(x-12)`

`y-1=(-4)/(-16)(x-12)`

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

Now standard form of equation of line =
`Ax+By=C`

So,
`y-1=(1)/(4)(x-12)`

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

`4y-4=(x-12)`

`12-4=x-4y`

`8=x-4y`

`x-4y=8`

Hence the standard form of the equation for this line is
`x-4y=8`