question_answer

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:

Here \[P(x,y)\] divides line segment AB

such that \[AP=\frac{3}{7}AB\]

\[\Rightarrow \frac{AP}{AB}=\frac{3}{7}\]

\[\Rightarrow \frac{AP}{AB}=\frac{7}{3}\]

\[\Rightarrow \frac{AB}{AP}-1=\frac{7}{3}-1\]

\[\Rightarrow \frac{AB-AP}{AP}=\frac{4}{3}\]

\[\Rightarrow \frac{BP}{AP}=\frac{4}{3}\]

\[\Rightarrow \frac{AP}{BP}=\frac{3}{4}\]

\[\therefore \] P divides AB in the ratio \[3:4\text{ (}m:n)\]

The coordinates of P are \[(x,y)\]

Therefore,

\[x=\frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},y=\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\]

\[x=\frac{3\times 2+4(-2)}{3+4},y=\frac{3(-4)+4(-2)}{3+4}\]

\[x=\frac{6-8}{7},y=\frac{-12-8}{7}\]

\[x=\frac{-2}{7},y=\frac{-20}{7}\]

Therefore, co-ordinates of \[P(x,y)\] are \[\left( \frac{-2}{7},\frac{-20}{7} \right)\]