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

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?\no x - 4y = 8\no x - 4y = 2\no 4x - y = 8\no 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?\no x - 4y = 8\no x - 4y = 2\no 4x - y = 8\no 4x - y = 2

Answer

Explanation:

Step1: Expand the point - slope form

Starting with $y - 1=\frac{1}{4}(x - 12)$, we distribute $\frac{1}{4}$ on the right - hand side: $y-1=\frac{1}{4}x - 3$.

Step2: Move all terms to one side

Subtract $\frac{1}{4}x$ from both sides and add 1 to both sides: $-\frac{1}{4}x+y=-3 + 1$. Simplify to get $-\frac{1}{4}x+y=-2$.

Step3: Eliminate the fraction

Multiply the entire equation by - 4 to get the standard form $Ax+By = C$ (where $A$, $B$, and $C$ are integers and $A\geq0$): $x-4y = 8$.

Answer:

A. $x - 4y=8$