Treat the coordinates, x from 1 to 4 and y from 1 to 3, separately. "Three times as far from A as from B". Think of that as "3" (distance from A) and "1" (distance from B). 3+ 1= 4 so divide the distance from 1 to 4 into 4 separate parts. 4-1= 3 so each part has x-length 3/4. The x-distance from A to the new point is 3(3/4)= 9/4 so the x-coordinate is 1+ 9/4= 13/4 or 3 and 1/4. Add another 3/4 to that and we get 4, the x-coordinate of B. The 3- 1= 2 so each part has y-length 2/4= 1/3. 3- 1/2= 5/2= 2 and 1/2 (subtract from because the point we want isbetweenA and B). The y-coordinate of the point is 1/2 and the point itself is (9/4, 5/2).

I said "the point we want is between A and B" because the problem specifiacally said "thepoint on the line. Actually, there is another point on the line through A and B and three times as from from A as from B. It is a pointbeyondB. 3- 1= 2 so divide the segement from A to B inhalf. The midpoint is ((1+ 4)/2, (1+ 3)/2)= (5/2, 2), an x-distance of 5/2- 1= 3/2 and a y-distance of 2- 1= 1. Multiplying each of those by 3, (1, 1)+ (9/2, 3)= (11/2, 4) lies on the line through A and B. The distance from A to that point is while the distance from B to that point is .