Indicate in standard form the equation of the line through the given points.

p(0, -4), q(5, 1)

Standard form is ax+by=c

First find the slope, (y2-y1)/(x2-x1)=(1--4)/(5-0)=5/5=1, so if we use the form:

y=mx+b we have

y=x+b, using either point, I'll use (5,1) we can solve for b

1=5+b

b=-4 so in slope intercept form we have:

y=x-4  Now we rearrange so that x and y are on one side and the constant is on the other side so that we have the standard form.  Note that by convention the coefficient of x should be positive so:

y-x=-4  multiplying the entire equation by -1 we get:

x-y=4

I found slope: 1--4/5-0 and got 1
Then I plugged the points (0,-4) into the slope equation
and ended up with y = x-4, when rewritten into standard form would be:
y-x = -4

hope it is correct if not i'm srry i tried to