If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = \(\frac{3}{7} \) AB and P lies on the line segment AB.

AP = \(\frac{3}{7} \) AB
7 AP = 3 AB
AB : AP = 7: 3
Let AB = 7x
\(\therefore \) AP = 3x
AB = AP + BP
7x = 3x + BP
BP = 7x - 3x = 4x

\(\therefore \frac{AP}{BP} = \frac{3x}{4x} \)
\(\therefore \) AP : BP = 3:4
by section formula ,

(x, y ) = \( (\frac{m1x2 + m2x1}{m1+m2} , \frac{m1y2 + m2y1}{m1+m2} ) \)
(x, y) = \( (\frac{3×2 +4×(-2) }{3+4} , \frac{3×(-4)+ 4(-2)}{3+4} ) \)
(x ,y ) = \( (\frac{6 -8 }{7} , \frac{-12+ -8}{7} ) \)
(x ,y ) = \( (\frac{-2}{7} , \frac{-20}{7} ) \)

Hence the coordinates of P = \( (\frac{-2}{7} , \frac{-20}{7} ) \)