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 $AB$.
Given:
$A$ and $B$ are two points having coordinates $(-2, -2)$ and $(2, -4)$ respectively.
To do:
We have to find the coordinates of $P$ such that $AP = \frac{3}{7} AB$.
Solution:
Let the coordinates of $P$ be $(x,y)$.
$PB = (1-\frac{3}{7})$
$AB=\frac{4}{7}AB$.
This implies,
$AP:PB=\frac{3}{7}AB:\frac{4}{7}AB=3:4$
Point $P$ divides the line segment joining the points $A(-2, -2)$ and $B(2, -4)$ in the ratio of $3 : 4$.
Using section formula, we have,
$(x, y)=(\frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n})$
Therefore,
$P(x,y)=(\frac{3 \times 2+4 \times (-2)}{3+4}, \frac{3 \times (-4)+4 \times (-2)}{3+4})$
$=(\frac{6-8}{7}, \frac{-12-8}{7})$
$=(\frac{-2}{7}, \frac{-20}{7})$
Therefore, the coordinates of $P$ are $(\frac{-2}{7}, \frac{-20}{7})$.
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