Answer:
P(0, 1)
Step-by-step explanation:
Using the section formula
[tex]x_{P}[/tex] = [tex]\frac{(3(-2))+(2(3))}{2+3}[/tex] = [tex]\frac{-6+6}{5}[/tex] = 0
[tex]y_{P}[/tex] = [tex]\frac{(3(5))+(2(-5))}{2+3}[/tex] = [tex]\frac{15-10}{5}[/tex] = 1
Hence P(0, 1)