Line jk passes through points j(–3, 11) and k(1, –3). what is the equation of line jk in standard form?

7x + 2y = –1
7x + 2y = 1
14x + 4y = –1
14x + 4y = 1

answer:

the first one 7x+2y=-1

step-by-step explanation:

i knows the math

I'll start with the point slope form and then rearrange for the standard

first find the slope of the line, m, which is equal to (dy/dx)=(-3-11)/(1--3)=-14/4

so m=-7/2, now use either point to solve for the y-intercept, or "b", in the line:

y=mx+b and we know -7/2 so

y=-7x/2 +b, and i'll use (1, -3) to solve for b:

-3=-7(1)/2+b

-6=-7+2b

b=1/2 so the line is:

y=(-7x+1)/2   now rearrange for x and y to be on one side and a constant on the other

2y=-7x+1

7x+2y=1

20y5 times 2 is 1010 times -2 is -20, plus y