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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?

Respuesta :

pluggedchocolate
      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:

[tex]x-4y=8[/tex]

Step-by-step explanation:

Given : (–4, –3) , (12, 1)

Solution:

[tex](x_1,y_1)=(12,1)[/tex]

[tex](x_2,y_2)=(-4,-3)[/tex]

Two point slope form : [tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]

Substitute the given values.

[tex]y-1=\frac{-3-1}{-4-12}(x-12)[/tex]

[tex]y-1=\frac{-4}{-16}(x-12)[/tex]

[tex]y-1=\frac{1}{4}(x-12)[/tex]

Now standard form of equation of line = [tex]Ax+By=C[/tex]

So, [tex]y-1=\frac{1}{4}(x-12)[/tex]

[tex]4(y-1)=(x-12)[/tex]

[tex]4y-4=(x-12)[/tex]

[tex]12-4=x-4y[/tex]

[tex]8=x-4y[/tex]

[tex]x-4y=8[/tex]

Hence  the standard form of the equation for this line is [tex]x-4y=8[/tex]

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