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

Answer:

$x-4y=8$

Step-by-step explanation:

Given : (–4, –3) , (12, 1)

Solution:

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

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

Two point slope form : $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Substitute the given values.

$y-1=\frac{-3-1}{-4-12}(x-12)$

$y-1=\frac{-4}{-16}(x-12)$

$y-1=\frac{1}{4}(x-12)$

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

So, $y-1=\frac{1}{4}(x-12)$

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

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

$12-4=x-4y$

$8=x-4y$

$x-4y=8$

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

Answer 2

      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