Given:
A has coordinates (-6,5)
C has coordinates (3.6,-0.4).
C divides AB in the ratio 3:2.
Refer to the diagram shown below.
The coordinates of B are determined by
[tex] \frac{3.6-(-6)}{x-(-6)} = \frac{3}{3+2} \\ \frac{9.6}{x+6} = \frac{3}{5} \\ 3(x+6) = 48 \\ 3x +18=48 \\ x=30/3=10[/tex]
Also,
[tex] \frac{5-(-0.4)}{5-y} = \frac{3}{5} \\ 3(5-y) = 27 \\ 15-3y=27 \\ y= \frac{12}{-3} =-4[/tex]
Answer:
The coordinates of B are (10,-4).
Now, we know the coordinates of line segment AB as A (-6, 5) and B (10,-4).
D (x,y) divides AB in the ratio 4:5.
Therefore
[tex] \frac{x-(-6)}{10-(-6)} = \frac{4}{4+5} \\ \frac{x+6}{16} = \frac{4}{9} \\ 9(x+6)=64 \\ 9x+54=64 \\ x=1.11[/tex]
Also,
[tex] \frac{5-y}{5-(-4)} = \frac{4}{9} \\ 5-y = 4 \\ y = 1[/tex]
Answer:
The coordinates of D are (1.11, 1)
