First of all, you need to recognize the realtionship between the areas of the triangles and the line segment lengths. The area of a right triangle (which these are) is half the product of the leg lengths. Here, that means
area ∆ADC : area ∆BDC = 3 : 4
(1/2)*DC*AC : (1/2)*DC*BC = 3 : 4 . . . . . substituting formulas for area
AC : BC = 3 : 4 . . . . . . . . . . . . . . . . . . . .. removing common factors
Now, it is a simple problem of dividing the line segment into the given ratio. This is what you've been doing all along. Here, I'll show it using the weighted average of the endpoint coordinates.
C = (4*A +3*B)/(4+3)
C = (4*(1, -9) +3*(8, -2)/7
C = (4*1 +3*8, 4*(-9) +3(-2))/7
C = (28, -42)/7
C = (4, -6)
The 4th selection is appropriate.
• (4,
-6)