Question

Find the coordinates of the points which divide, internally and externally, the line joining the point $(a+b,a-b)$ to the point $(a-b,a+b)$ in the ratio $a:b$.

Answer 1

The coordinates of point when ${(x}_1,y_1)$ and ${(x}_2,y_2)$ are divided in $m:n$

$\left(i\right)$ internally is $(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

$\left(ii\right)$ externally $(\frac{m{x}_2-n{x}_1}{m-n},\frac{m{y}_2-n{y}_1}{m-n})$

Let the point be $P(h,k)$

(1) Internal division

$h=\frac{a(a-b)+b(a+b)}{a+b}=\frac{a^2-ab+ab+b^2}{a+b}=\frac{a^2+b^2}{a+b}k=\frac{a(a+b)+b(a-b)}{a+b}=\frac{a^2+ab+ab-b^2}{a+b}=\frac{a^2+2ab-b^2}{a+b}\Rightarrow P(\frac{a^2+b^2}{a+b},\frac{a^2+2ab-b^2}{a+b})$

(2) External division

$h=\frac{a(a-b)-b(a+b)}{a-b}=\frac{a^2-ab-ab-b^2}{a-b}=\frac{a^2-2ab-b^2}{a-b}k=\frac{a(a+b)-b(a-b)}{a-b}=\frac{a^2+ab-ab+b^2}{a-b}=\frac{a^2+b^2}{a-b}\Rightarrow P(\frac{a^2-2ab-b^2}{a-b},\frac{a^2+b^2}{a-b})$