Perpendicular bisector of a segment

Statement

Write the perpendicular bisector of the segment $AB$ with $A(1;3)$, $B(-3;5)$ and verify that it is perpendicular to $AB$ and passes through the midpoint $M(-1;4)$.

Solution

A point $P(x;y)$ belongs to the perpendicular bisector when $d(P,A) = d(P,B)$: $\sqrt{(x-1)^2 + (y-3)^2} = \sqrt{(x+3)^2 + (y-5)^2}.$ Squaring (both sides are distances, hence $\ge 0$) and expanding: $x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 + 6x + 9 + y^2 - 10y + 25.$ The terms $x^2$ and $y^2$ cancel; what remains is $-2x - 6y + 10 = 6x - 10y + 34$, that is $8x - 4y + 24 = 0$, namely $\boxed{2x - y + 6 = 0} \quad\Longleftrightarrow\quad y = 2x + 6.$ Check. The gradient of the perpendicular bisector is $2$; that of $AB$ is $m_{AB} = \dfrac{5-3}{-3-1} = -\dfrac{1}{2}$. Since $2 \cdot \left(-\tfrac{1}{2}\right) = -1$, the perpendicular bisector is perpendicular to $AB$. Moreover at $M(-1;4)$: $2(-1) - 4 + 6 = 0$, so the perpendicular bisector passes through the midpoint. ✓

Links

Topics: Geometric loci
Concepts: Perpendicular bisector of a segment · Distance between points · Equation of a line · Cartesian plane · Line
Skills: Analytical geometry · Solving equations
Exercise type: Geometric problem