General case
For P₁=(−2,1) and P₂=(4,5), Δx=6, Δy=4, and M=(1,3). The point-normal equation is 6(x−1)+4(y−3)=0, which simplifies to 3x+2y−9=0 or y=−3x/2+9/2.
Enter two distinct points to find the line that crosses their midpoint at 90°. The calculator returns exact and decimal forms, a graph, and substituted steps.
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For P₁=(x₁,y₁) and P₂=(x₂,y₂), the midpoint is:
M=(h,k)=((x₁+x₂)/2,(y₁+y₂)/2)
The segment direction is (Δx,Δy)=(x₂−x₁,y₂−y₁). This becomes a normal vector to the perpendicular bisector.
Use the point-normal equation through the midpoint:
Δx(x−h)+Δy(y−k)=0
This vector form works without special cases. If the original segment is horizontal, it simplifies to the vertical line x=h. If the segment is vertical, it simplifies to the horizontal line y=k.
The bisector must pass through the midpoint. It must also be perpendicular to the segment, so the segment direction (Δx,Δy) is a normal vector to the bisector. A point (x,y) lies on that line when the dot product (Δx,Δy)·(x−h,y−k) is zero.
For P₁=(−2,1) and P₂=(4,5), Δx=6, Δy=4, and M=(1,3). The point-normal equation is 6(x−1)+4(y−3)=0, which simplifies to 3x+2y−9=0 or y=−3x/2+9/2.
For P₁=(−3,2) and P₂=(5,2), the midpoint is (1,2). The segment is horizontal, so its perpendicular bisector is the vertical line x=1.
For P₁=(4,−2) and P₂=(4,6), the midpoint is (4,2). The segment is vertical, so its perpendicular bisector is the horizontal line y=2.
For P₁=(1/2,−1) and P₂=(7/2,3), the midpoint is (2,1). Here Δx=3 and Δy=4, so the bisector is 3(x−2)+4(y−1)=0, or 3x+4y−10=0.
Ax+By+C=0 is the most reliable all-purpose form. It represents horizontal and vertical lines without exceptions.
A(x−h)+B(y−k)=0 shows both the midpoint and the segment direction used as the line's normal vector.
y=mx+b is convenient for graphing nonvertical lines. A vertical bisector is reported as x=h instead.
Every point on the bisector is equally far from the two endpoints. This is why triangle perpendicular bisectors meet at the circumcenter.
Find the midpoint (h,k), calculate Δx=x₂−x₁ and Δy=y₂−y₁, then use Δx(x−h)+Δy(y−k)=0.
For endpoints (x₁,y₁) and (x₂,y₂), the formula is (x₂−x₁)(x−(x₁+x₂)/2)+(y₂−y₁)(y−(y₁+y₂)/2)=0.
A horizontal segment has the vertical bisector x=h. A vertical segment has the horizontal bisector y=k, where (h,k) is the midpoint.
Identical points have no segment direction. Infinitely many lines pass through that one location, so no unique perpendicular bisector exists.
Yes. Conversely, every point equidistant from two distinct endpoints lies on their perpendicular bisector.
Yes. Examples include -3/4, -2 1/3, 1.25, and 4e-3. Rational results remain exact.
Substitute the midpoint into the equation; the left side should equal zero. A direction vector along the bisector should also have dot product zero with the original segment direction.
Calculation method: integers, finite decimals, scientific notation, fractions, and mixed numbers are converted to reduced rational numbers. Midpoints and line coefficients are calculated with exact integer arithmetic. Standard-form coefficients are cleared of denominators, divided by their greatest common divisor, and sign-normalized. Only graph coordinates, segment length, and requested decimal approximations use floating-point arithmetic.
Formula references: the midpoint method follows OpenStax College Algebra; the geometric definition and equal-distance property follow Wolfram MathWorld's perpendicular bisector reference.
Verification completed July 30, 2026. Exact-arithmetic checks cover general, horizontal, vertical, fractional, negative, identical-point, empty, malformed, and large-coordinate cases.