DE : DF = 2 : 5
So, DE = 2 parts and DF = 5 parts
EF = DF - DE
= 5 - 2
= 3 parts
DE : EF = 2 : 3
Let D be (x, y).
Then, the point E divides the line segment DF internally in the ratio 2 : 3.
So, E is [tex](\frac{2(-3)+3x}{2+3}, \frac{2(-4)+3y}{2+3} )[/tex]
[tex]=(\frac{3x-6}{5}, \frac{3y-8}{5} )[/tex]
But, it is given that E is (-3, 2).
Therefore,
[tex]\frac{3x-6}{5} = -3 and \frac{3y-8}{5}=2[/tex]
3x - 6 = -15 and 3y - 8 = 10
3x = -9 and 3y = 18
x = -3 and y = 6
Hence, D is (-3, 6)
