the point - slope form of the equation of the line that passes through (-4, -3) and (12, 1) is ( y…

the point - slope form of the equation of the line that passes through (-4, -3) and (12, 1) is ( y - 1=\frac{1}{4}(x - 12) ). what is the standard form of the equation for this line?\n( \bigcirc x - 4y = 8 )\n( \bigcirc x - 4y = 2 )\n( \bigcirc 4x - y = 8 )\n( \bigcirc 4x - y = 2 )

the point - slope form of the equation of the line that passes through (-4, -3) and (12, 1) is ( y - 1=\frac{1}{4}(x - 12) ). what is the standard form of the equation for this line?\n( \bigcirc x - 4y = 8 )\n( \bigcirc x - 4y = 2 )\n( \bigcirc 4x - y = 8 )\n( \bigcirc 4x - y = 2 )

Answer

Explanation:

Step1: Expand the point - slope form

Multiply both sides of (y - 1=\frac{1}{4}(x - 12)) by (4) to get (4(y - 1)=x - 12). Then expand: (4y-4=x - 12).

Step2: Rearrange to standard form (Ax + By=C)

Move (x) and constants: (-x+4y=- 12 + 4), which simplifies to (-x + 4y=-8). Multiply through by (-1) to make (x) coefficient positive: (x-4y = 8).

Answer:

A. (x - 4y=8)