What is the Equation of a Perpendicular Bisector?

In geometry, a perpendicular bisector (đường trung trực in Vietnamese) is a line that divides a line segment into two equal parts at a 90-degree angle. Let’s delve into how to find its equation.

Key Concepts

Line Segment and Midpoint

Consider a line segment with endpoints $A(x_1, y_1)$ and $B(x_2, y_2)$. The midpoint $M$ of this segment is calculated as:
$M bigg(frac{x_1 + x_2}{2}, frac{y_1 + y_2}{2}bigg)$

Slope of the Line Segment

The slope $m$ of the line segment $AB$ is given by:
$m = frac{y_2 – y_1}{x_2 – x_1}$

Slope of the Perpendicular Bisector

The slope of the perpendicular bisector is the negative reciprocal of the slope of $AB$. If the slope of $AB$ is $m$, then the slope of the perpendicular bisector is $-frac{1}{m}$

Deriving the Equation

To find the equation of the perpendicular bisector, we use the point-slope form of a line equation:
$y – y_1 = m(x – x_1)$

Substituting the midpoint $M$ and the slope $-frac{1}{m}$, we get:
$y – frac{y_1 + y_2}{2} = -frac{1}{frac{y_2 – y_1}{x_2 – x_1}} bigg(x – frac{x_1 + x_2}{2}bigg)$

Simplifying, we obtain:
$y – frac{y_1 + y_2}{2} = -frac{x_2 – x_1}{y_2 – y_1} bigg(x – frac{x_1 + x_2}{2}bigg)$

Example

Let’s consider an example where the endpoints of the line segment are $A(2, 3)$ and $B(4, 7)$

  1. Midpoint Calculation:
    $M bigg(frac{2 + 4}{2}, frac{3 + 7}{2}bigg) = (3, 5)$

  2. Slope of $AB$:
    $m = frac{7 – 3}{4 – 2} = 2$

  3. Slope of Perpendicular Bisector:
    $-frac{1}{2}$

  4. Equation of Perpendicular Bisector:
    Using the point-slope form:
    $y – 5 = -frac{1}{2}(x – 3)$
    Simplifying:
    $y – 5 = -frac{1}{2}x + frac{3}{2}$
    $y = -frac{1}{2}x + frac{13}{2}$

Conclusion

Understanding the equation of the perpendicular bisector involves calculating the midpoint and the slope of the original line segment. This knowledge is crucial for various applications in geometry, such as constructing perpendicular bisectors in triangle circumcircles.

Citations

  1. 1. Khan Academy – Perpendicular Bisectors
  2. 2. Math Open Reference – Perpendicular Bisector
  3. 3. Purplemath – Perpendicular Bisectors

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ