## Task 3 - 030933

Which points $P(x ; 0)$ are twice as far from the point $P_{1}(a ; 0)$ as from $P_{2}(b ; 0)$?

Determine the abscissas of these points! $(b>a)$
Given: $\left|P_{1} P\right|=2 \cdot\left|P_{2} P\right|$ and $|x-a|=2 \cdot|x-b|$

1. Case: $x \leq a<b$ does not occur, as $P$ should be farther from $P_{1}$ than from $P_{2}$.
2. Case: $a \leq x \leq b: x-a=2 \cdot(b-x) \Rightarrow x=a+(b-a) \cdot \frac{2}{3}$
3. Case: $a<b \leq x: x-a=2 \cdot(x-b) \Rightarrow x=2 \cdot b-a$

The points $\left(a+\frac{2}{3}(b-a) ; 0\right)$ and $(2 \cdot b-a ; 0)$ are twice as far from the point $P_{1}(a ; 0)$ as from $P_{2}(b ; 0)$.