The point slope form of the equation of the line that passes through (-4,3) and (12,1) is y-1=1\4(x-12) what is the standard form of the equation for this line

Something is wrong here, the slope is off...

m=(y2-y1)/(x2-x1)

m=(1-3)/(12--4)

m=-2/16

m=-1/8

So the point slope form should be:

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

The standard form of the line is ax+by=c so we need to rearrange it to reflect the this form...multiply both sides by -8 to get:

8-8y=x-12  subtract 8 from both sides

-8y=x-20  subtract x from both sides

-x-8y=-20  by convention, the coefficient of x should be positive, multiply both sides by -1

x+8y=20



Just to show that the equation given you is incorrect..

y-1=(x-12)/4  using the point (-4,3)

3-1=(-4-12)/4

2=-16/4

2=-4  which is obviously incorrect...(they had the wrong slope for the two points given)