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

You can start with the form
  ∆y(x -x1) -∆x(y -y1) = 0
Here, we have
  ∆y = 11-(-3) = 14
  ∆x = -3-1 = -4
and we can choose (x1, y1) = (1, -3). This gives
  14(x -1) -(-4)(y -(-3)) = 0
  14x +4y -2 = 0
All these terms have a common factor of 2 that we can remove. Adding 1 to the result puts it in standard form:

  7x +2y = 1