If the coordinates of points A and B are (−2, −2) and (2, −4) respectively,

The coordinates of the points $A$ and $B$ are $(-2,-2)$ and $(2,-4)$ respectively, where $A P=\frac{3}{7} A B$ and $P$ lies on the line segment $A B$. So

$A P+B P=A B$

$\Rightarrow A P+B P=\frac{7 A P}{3} \quad \because A P=\frac{3}{7} A B$

$\Rightarrow B P=\frac{7 A P}{3}-A P$

$\Rightarrow \frac{A P}{B P}=\frac{3}{4}$

Let (x, y) be the coordinates of P which divides AB in the ratio 3 : 4 internally. Then

$x=\frac{3 \times 2+4 \times(-2)}{3+4}=\frac{6-8}{7}=-\frac{2}{7}$

$y=\frac{3 \times(-4)+4 \times(-2)}{3+4}=\frac{-12-8}{7}=-\frac{20}{7}$

Hence, the coordinates of point $P$ are $\left(-\frac{2}{7},-\frac{20}{7}\right)$.