1. **Stating the problem:** We want to understand why the condition for a vector $w$ to be equidistant from vectors $u$ and $v$ is given by $$ (w - u) \cdot (w - u) = (w - v) \cdot (w - v). $$ Then, we want to show this condition is equivalent to $$ \left(w - \frac{1}{2}(u + v)\right) \cdot (u - v) = 0 $$ and explain why this means $w$ lies on the perpendicular bisector of the segment joining $u$ and $v$. 2. **Formula and explanation:** The squared distance between two vectors $a$ and $b$ is given by the dot product $$ (a - b) \cdot (a - b). $$ So, $w$ is equidistant from $u$ and $v$ means $$ |w - u| = |w - v| \implies (w - u) \cdot (w - u) = (w - v) \cdot (w - v). $$ This equality states that the squared lengths of the vectors from $w$ to $u$ and $w$ to $v$ are the same. 3. **Showing equivalence:** Start with the given condition: $$ (w - u) \cdot (w - u) = (w - v) \cdot (w - v). $$ Expand both sides: $$ w \cdot w - 2 w \cdot u + u \cdot u = w \cdot w - 2 w \cdot v + v \cdot v. $$ Cancel $w \cdot w$ from both sides: $$ - 2 w \cdot u + u \cdot u = - 2 w \cdot v + v \cdot v. $$ Rearranged: $$ 2 w \cdot v - 2 w \cdot u = v \cdot v - u \cdot u. $$ Factor left side: $$ 2 w \cdot (v - u) = v \cdot v - u \cdot u. $$ Divide both sides by 2: $$ w \cdot (v - u) = \frac{v \cdot v - u \cdot u}{2}. $$ Rewrite the right side using the identity: $$ v \cdot v - u \cdot u = (v - u) \cdot (v + u). $$ So, $$ w \cdot (v - u) = \frac{(v - u) \cdot (v + u)}{2} = (v - u) \cdot \frac{v + u}{2}. $$ Bring all terms to one side: $$ w \cdot (v - u) - (v - u) \cdot \frac{v + u}{2} = 0. $$ Since dot product is commutative, $$ (w - \frac{u + v}{2}) \cdot (v - u) = 0. $$ Because $v - u = -(u - v)$, this is equivalent to $$ \left(w - \frac{u + v}{2}\right) \cdot (u - v) = 0. $$ 4. **Interpretation:** The vector $\frac{u + v}{2}$ is the midpoint of the segment joining $u$ and $v$. The equation $$ \left(w - \frac{u + v}{2}\right) \cdot (u - v) = 0 $$ states that the vector from the midpoint to $w$ is orthogonal (perpendicular) to the vector $u - v$. This means $w$ lies on the line perpendicular to the segment $uv$ at its midpoint — the perpendicular bisector. **Final answer:** The condition for $w$ to be equidistant from $u$ and $v$ is that $w$ lies on the perpendicular bisector of the segment joining $u$ and $v$, expressed by $$ \left(w - \frac{1}{2}(u + v)\right) \cdot (u - v) = 0. $$