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Finding Internal and External Division Points on a Line 📂Geometry

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

5964FCC52.png

$$\overline{AP}:\overline{PB} = m:n$$

$$\implies x-x_{1}:x_{2}-x=m:n$$

$$\implies mx_{2}-mx=nx-nx_{1}$$

$$\implies mx+nx=mx_{2}+nx_{1}$$

$$\implies (m+n)x=mx_{2}+nx_{1}$$

$$\implies x=\frac{mx_{2}+nx_{1}}{m+n}$$

External Division Point

5964FCC62.png

$$\overline{AQ}:\overline{BQ} = m:n$$

$$\implies x-x_{1}:x-x_{2}=m:n$$

$$\implies mx-mx_{2}=nx-nx_{1}$$

$$\implies mx-nx=mx_{2}-nx_{1}$$

$$\implies (m-n)x=mx_{2}-nx_{1}$$

$$\implies x=\frac{mx_{2}-nx_{1}}{m-n}$$