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 standa the equation for this line?

Answer 1

Answer:

The equation would be -x + 4y = -8

Step-by-step explanation:

To find the standard form of the line, solve for the constant.

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

y - 1 = 1/4x - 3

-1/4x + y - 1 = -3

-1/4x + y = -2

-x + 4y = -8

Answer 2

Answer: The standard equation of this line is x-4y=8.

Step-by-step explanation:

Since we have given that

Two coordinates are as follows:

(-4,-3) and (12,1)

So, slope would be

$\dfrac{y_2-y_1}{x_2-x_1}\\\\\\=\dfrac{1-(-3)}{12-(-4)}\\\\\\=\dfrac{1+3}{12+4}\\\\\\=\dfrac{4}{16}\\\\\\=\dfrac{1}{4}$

So, the standard form of equation of this line is given by

$y-y_1=\dfrac{1}{4}(x-x_1)\\\\y-(-3)=\dfrac{1}{4}(x-(-4)\\\\y+3=\dfrac{1}{4}(x+4)\\\\4(y+3)=x+4\\\\4y+12=x+4\\\\4y=x+4-12\\\\4y=x-8\\\\-x+4y=-8\\\\x-4y=8$

Hence, the standard equation of this line is x-4y=8.