Find the point, m, that divides segment ab into a ratio of 5: 2 if a is at (1, 2) and b is at (8, 16). a) (6, 12) b) (-6, 12) c) (6, -12) d) (-6, -12)

Since M divides segment AB into a ratio of 5:2, we can say that M is 5/(5+2) of the length of AB. Therefore 5/7 × AB.
distance of AB = d
5/7×(x2 - x1) for the x and 5/7×(y2 - y1) for the y
5/7×(8 - 1) = 5/7 (7) = 5 for the x
and 5/7×(16 - 2) = 5/7 (14) = 10 for the y
But remember the line AB starts at A (1, 2),
so add 1 to the x: 5+1 = 6
and add 2 to the y: 10+2 = 12
Therefore the point M lies exactly at...
A)(6, 12)