Question

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

Answer 1

Given coordinates of point $A(-2,-2)$ and $B(2,-4)$ and point $P$ divided $AB$ as $AP=\frac{3}{7}AB$

Then $BP=\frac{4}{7}AB$

So point $P$ divided $AB$ in ratio $3:4$

$m=3$ and $n=4$

Using Section formula, coordinates of point $P$ are

$[x=\left(\frac{m{x}_1+n{x}_2}{m+n}\right)]$ and $[y=\left(\frac{m{y}_1+n{y}_2}{m+n}\right)]$

$A(-2,-2)\equiv (x_2,y_2)$ and $B(2,-4)\equiv (x_1,y_1)$

$P=(\frac{3\times 2-2\times 4}{3+4},\frac{-4\times 32\times 4}{3+4})=(\frac{6-8}{7},\frac{-12-8}{7})=(-\frac{2}{7},-\frac{20}{7})$