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? x – 4y = 8 x – 4y = 2 4x – y = 8 4x – y = 2

Answer 1

Answer:  The correct option is

(A) $x-4y=8.$

Step-by-step explanation:  Given that the point-slope form of the equation of the line passing through the points (-4, -3) and (12, 1) is given by

$y-1=\dfrac{1}{4}(x-12)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We are to select the correct standard form of the equation for the above line.

From equation (i), we have

$y-1=\dfrac{1}{4}(x-12)\\\\\\\Rightarrow 4(y-1)=x-12\\\\\Rightarrow 4y-4=x-12\\\\\Rightarrow x-12-4y+4=0\\\\\Rightarrow x-4y-8=0\\\\\Rightarrow x-4y=8.$

Thus, the required standard form of the equation of the line is

$x-4y=8.$

Option (A) is CORRECT.

Answer 2

(y - 1) = 1/4(x - 12)
4(y - 1) = x - 12
4y - 4 = x - 12
x - 4y = -4 + 12
x - 4y = 8