x+y=k => y=k-x
\(\dfrac{x-a} {m}=\dfrac{y-b} {n}=>\dfrac{x-a}{m} = \dfrac{k-x-b} {n}=> 2x=\dfrac{a} {m}+k-\dfrac{b} {n}=>x=(\dfrac{a} {m}+k-\dfrac{b} {n})/2 => y=k-x=(k-\dfrac { a} {m}+\dfrac{b} {n}/2 \)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{1x-a}{m}=\dfrac{y-b}{n}=\dfrac{x+y-a-b}{m+n}=\dfrac{k-a-b}{m+n}\)
\(1\Rightarrow\left\{{}\begin{matrix}\dfrac{x-a}{m}=\dfrac{k-a-b}{m+n}\\\dfrac{y-b}{n}=\dfrac{k-a-b}{m+n}\end{matrix}\right.\)
\(1\Rightarrow\left\{{}\begin{matrix}x-a=\dfrac{m\left(k-a-b\right)}{m+n}\\y-b=\dfrac{n\left(k-a-b\right)}{m+n}\end{matrix}\right.\)
\(1\Rightarrow\left\{{}\begin{matrix}x=a+\dfrac{m\left(k-a-b\right)}{m+n}\\y=b+\dfrac{n\left(k-a-b\right)}{m+n}\end{matrix}\right.\)
