Let u=pq be the directed line segment

The vector u is <9, 12> which means "start at any point, go to the right 9 units, and then up 12 units". If the starting point is point P = (0,0), then the terminal point is in quadrant I. For vector cu, the terminal point is now in quadrant III because we multiply the coordinates of (9,12) by some negative number c (eg: c = -1 so the terminal point is (-9,-12))

Answer: Choice B) terminal point of vector cu is in quadrant 3