Graph the line containing the point p and having slop m
Lines with a slope of zero (m = 0) are horizontal, (i.e. flat, parallel to the x-axis). Now, if one point on the line is 4 units below the x-axis, (i.e. -4 units above the x-axis), then every point on the line is 4 units below the x-axis. So it's a horizontal line 4 units below the x-axis.
Lines have the form y = mx + b, we know m = 0 so, y = 0x + b = b, (because 0x = 0, and 0 + b = b ). So this line has the form y = b. We know a point on the line is (2,-4), which means x = 2, y = -4 satisfies the equation. Now, plug -4 into our equation for y and you get -4 = b, or b = -4. Substitute -4 for b back into the y = b equation and we simply get y = -4. y = -4 is the equation for this line, the line with slope m = 0, and passing trough the point (2,-4).
The big clue to this question is looking at the slope of the line, which is 0. This means that the line is completely horizontal (parallel to the x axis) and that the "y" values of the line does not change. As a result, every point on this line will have a y value of -4. Other points on this line, in addition to P, are (1,-4; 5,-4, ect). You only need two points to graph a line. Hope this helps!