Which of the following statements is correct about the pixels generated for…

Bresenham's line-drawing algorithm plots a raster line using only integer arithmetic. For a segment whose slope lies between 0 and 1 (here dx = x2 - x1 is greater than or equal to dy = y2 - y1, both positive), it advances x by exactly one pixel at every step and uses an integer decision parameter p to decide whether y also advances. Starting from p0 = 2*dy - dx: if p is negative, y is held at its current value and p is updated as p = p + 2*dy; if p is zero or positive, y is incremented by 1 and p is updated as p = p + 2*dy - 2*dx.

1. Compute dx = 25 - 20 = 5 and dy = 14 - 10 = 4, so the initial decision parameter is p0 = 2*dy - dx = 2(4) - 5 = 3.

2. Starting pixel: x = 20, y = 10 (the segment's first endpoint), with p0 = 3.

3. x = 21: since p0 = 3 is >= 0, y increments to 11, giving pixel (21, 11); update p1 = p0 + 2*dy - 2*dx = 3 + 8 - 10 = 1.

4. x = 22: since p1 = 1 is >= 0, y increments to 12, giving pixel (22, 12); update p2 = p1 + 2*dy - 2*dx = 1 + 8 - 10 = -1.

5. x = 23: since p2 = -1 is < 0, y stays at 12, giving pixel (23, 12); update p3 = p2 + 2*dy = -1 + 8 = 7.

6. x = 24: since p3 = 7 is >= 0, y increments to 13, giving pixel (24, 13); update p4 = p3 + 2*dy - 2*dx = 7 + 8 - 10 = 5.

7. x = 25: since p4 = 5 is >= 0, y increments to 14, giving pixel (25, 14), the segment's second endpoint.

Since dx >= dy, exactly one pixel is plotted per unit increase in x, so this segment must contain exactly dx + 1 = 6 pixels in total: (20,10), (21,11), (22,12), (23,12), (24,13), (25,14). This matches the trace above, rules out totals of seven or five, and confirms the true initial decision parameter is 2*dy - dx = 3, not 6.

The statement that holds for this segment is that the pixel (23, 12) is generated by the algorithm - confirmed directly at the x = 23 step of the trace above, where the decision parameter is negative and y is held at 12.