Finding Internal and External Division Points on a Line
Theorem
The coordinates of the point $A(x_{1})$ and the point $B(x_{2})$ on the number line, dividing internally at $m:n$, are $\displaystyle x=\frac{mx_{2}+nx_{1}}{m+n}$, and the coordinates of the point that divides externally at $m:n$ are $\displaystyle x=\frac{mx_{2}-nx_{1}}{m-n}$.
Explanation
By examining the formulas for the internal and external division point, one can see that only the signs are different. There is no need to memorize both expressions; simply memorize the internal division point formula and change the sign accordingly. In fact, deriving the formula is quick and easy, so there is no real need to memorize it either.
Proof
Internal Division Point
$$\overline{AP}:\overline{PB} = m:n$$
$$\implies x-x_{1}:x_{2}-x=m:n$$
$$\implies mx_{2}-mx=nx-nx_{1}$$
$$\implies mx+nx=mx_{2}+nx_{1}$$
$$\implies (m+n)x=mx_{2}+nx_{1}$$
$$\implies x=\frac{mx_{2}+nx_{1}}{m+n}$$
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External Division Point
$$\overline{AQ}:\overline{BQ} = m:n$$
$$\implies x-x_{1}:x-x_{2}=m:n$$
$$\implies mx-mx_{2}=nx-nx_{1}$$
$$\implies mx-nx=mx_{2}-nx_{1}$$
$$\implies (m-n)x=mx_{2}-nx_{1}$$
$$\implies x=\frac{mx_{2}-nx_{1}}{m-n}$$
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