krit.club logo

Geometry and Trigonometry - Perpendicular bisector of a line and Voronoi diagrams-extended

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

The perpendicular bisector of a line segment ABAB is a line that passes through the midpoint MM of ABAB at a 90∘90^{\circ} angle. Every point on this bisector is equidistant from the endpoints AA and BB.

A line segment AB with its perpendicular bisector passing through the midpoint M.
•

A Voronoi diagram partitions a plane into regions based on distance to specific points (sites). Each region consists of all points closer to its own site than to any other site. The boundaries (edges) of these regions are segments of perpendicular bisectors between adjacent sites.

A simple Voronoi diagram with three sites and three meeting edges.
•

A Voronoi vertex is the point where three or more Voronoi edges meet. This vertex is equidistant from the three nearest sites and is the center of a circumcircle passing through those sites.

•

To find the equation of a perpendicular bisector: 1. Calculate the midpoint MM of the segment. 2. Find the gradient mm of the segment. 3. Determine the perpendicular gradient m⊥=−1mm_{\perp} = -\frac{1}{m}. 4. Use the point-slope form with MM and m⊥m_{\perp}.

📐Formulae

M=(x1+x22,y1+y22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) (Midpoint Formula)

m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} (Gradient Formula)

m1×m2=−1  ⟹  m⊥=−1mm_1 \times m_2 = -1 \implies m_{\perp} = -\frac{1}{m} (Perpendicular Gradient)

y−y1=m(x−x1)y - y_1 = m(x - x_1) (Point-Slope Equation of a line)

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} (Distance Formula)

💡Examples

Problem 1:

Find the equation of the perpendicular bisector of the line segment joining the points A(2,4)A(2, 4) and B(6,10)B(6, 10).

Solution:

  1. Find the midpoint MM: M=(2+62,4+102)=(4,7)M = \left( \frac{2+6}{2}, \frac{4+10}{2} \right) = (4, 7)
  2. Find the gradient of ABAB: mAB=10−46−2=64=32m_{AB} = \frac{10-4}{6-2} = \frac{6}{4} = \frac{3}{2}
  3. Find the perpendicular gradient: m⊥=−13/2=−23m_{\perp} = -\frac{1}{3/2} = -\frac{2}{3}
  4. Use the point-slope form with M(4,7)M(4, 7): y−7=−23(x−4)y - 7 = -\frac{2}{3}(x - 4)
  5. Simplify: 3(y−7)=−2(x−4)  ⟹  3y−21=−2x+8  ⟹  2x+3y=293(y - 7) = -2(x - 4) \implies 3y - 21 = -2x + 8 \implies 2x + 3y = 29

Explanation:

To find the perpendicular bisector, we first determine the point it must pass through (the midpoint) and its slope (the negative reciprocal of the original line's slope).

Problem 2:

Three cell towers are located at S1(0,0)S_1(0, 0), S2(4,0)S_2(4, 0), and S3(0,6)S_3(0, 6). Find the coordinates of the Voronoi vertex formed by these three sites.

Solution:

  1. The Voronoi vertex is the intersection of the perpendicular bisectors.
  2. Perpendicular bisector of S1(0,0)S_1(0,0) and S2(4,0)S_2(4,0): The midpoint is (2,0)(2, 0) and the line is vertical because S1S2S_1S_2 is horizontal. Equation: x=2x = 2
  3. Perpendicular bisector of S1(0,0)S_1(0,0) and S3(0,6)S_3(0,6): The midpoint is (0,3)(0, 3) and the line is horizontal because S1S3S_1S_3 is vertical. Equation: y=3y = 3
  4. The intersection of x=2x = 2 and y=3y = 3 is the point (2,3)(2, 3).

Explanation:

The Voronoi vertex is the point equidistant from all three sites. Since two of the bisectors are simple horizontal and vertical lines, their intersection (2,3)(2, 3) is easily found and represents the vertex.

Problem 3:

Determine the equation of the perpendicular bisector of the line segment connecting P(−2,3)P(-2, 3) and Q(4,3)Q(4, 3).

Horizontal segment PQ and its vertical perpendicular bisector x=1.

Solution:

  1. Find the midpoint MM: M=(−2+42,3+32)=(1,3)M = \left( \frac{-2 + 4}{2}, \frac{3 + 3}{2} \right) = (1, 3)
  2. Find the gradient mm of PQPQ: m=3−34−(−2)=06=0m = \frac{3 - 3}{4 - (-2)} = \frac{0}{6} = 0
  3. Since the line PQPQ is horizontal (m=0m=0), the perpendicular bisector must be a vertical line passing through the x-coordinate of the midpoint.
  4. The equation is x=1x = 1.

Explanation:

Because the segment is horizontal, its perpendicular bisector is a vertical line. Vertical lines have the form x=kx = k, where kk is the x-coordinate of the midpoint.

Problem 4:

Two sites in a Voronoi diagram are located at A(2,1)A(2, 1) and B(2,5)B(2, 5). A third site is at C(6,3)C(6, 3). Find the equation of the Voronoi edge separating sites AA and CC.

The perpendicular bisector representing the Voronoi edge between sites A and C.

Solution:

  1. Midpoint MAC=(2+62,1+32)=(4,2)M_{AC} = \left( \frac{2+6}{2}, \frac{1+3}{2} \right) = (4, 2)
  2. Gradient mAC=3−16−2=24=0.5m_{AC} = \frac{3-1}{6-2} = \frac{2}{4} = 0.5
  3. Perpendicular gradient m⊥=−10.5=−2m_{\perp} = -\frac{1}{0.5} = -2
  4. Equation: y−2=−2(x−4)y - 2 = -2(x - 4) y−2=−2x+8y - 2 = -2x + 8 y=−2x+10y = -2x + 10

Explanation:

The Voronoi edge between two sites is the perpendicular bisector of the segment connecting them. We find the midpoint and the negative reciprocal of the gradient of ACAC to construct the line equation.