Question

A point moves so that is always equidistant from (-5,5) and (-2,2) . Find the equation of its locus

Answer 1

$\begin{aligned} &\text { The locus of a point equidistant from two fixed points is the perpendicular bisector of } \\ &\text { the line segment joining these two fixed points. } \\ &\text { In this case, the two fixed points are given as } A(-5,5) \text { and } B(-2,2) \text {. } \\ &\Rightarrow \quad \text { The midpoint of } A B \text { is }\left(-\frac{7}{2}, \frac{7}{2}\right) \text {. } \\ &\text { The slope of } A B \text { is } \frac{(5-2)}{(-5+2)}=-1 \text {. } \\ &\Rightarrow \quad \text { The slope of the line perpendicular to } A B \text { is } 1 \text {. } \\ &\Rightarrow \quad \text { The perpendicular bisector of } A B \text { passes through point }\left(-\frac{7}{2}, \frac{7}{2}\right) \text { and has a } \\ &\Rightarrow \quad \text { The equation of the perpendicular bisector of } A B \text { is } y-\frac{7}{2}=1\left(x+\frac{7}{2}\right) . \\ &\Rightarrow \quad \text { The equation of the perpendicular bisector of } A B \text { is } x-y+7=0 . \end{aligned}$