CK-12 Geometry Concepts

(Elliott) #1

5.2. Perpendicular Bisectors http://www.ck12.org


Practice


1.mis the perpendicular bisector ofAB.
a. List all the congruent segments.
b. IsConAB? Why or why not?
c. IsDonAB? Why or why not?

For Question 2, determine if


←→


STis the perpendicular bisector ofXY. Explain why or why not.

2.

For Questions 3-7, consider line segmentABwith endpointsA( 2 , 1 )andB( 6 , 3 ).



  1. Find the slope ofAB.

  2. Find the midpoint ofAB.

  3. Find the equation of the perpendicular bisector ofAB.

  4. FindAB. Simplify the radical, if needed.

  5. PlotA,B, and the perpendicular bisector. Label itm. How could you find a pointConm, such thatCwould
    be the third vertex of equilateral triangle 4 ABC?You do not have to find the coordinates, just describehow
    you would do it.


For Questions 8-12, consider 4 ABCwith verticesA( 7 , 6 ),B( 7 ,− 2 )andC( 0 , 5 ). Plot this triangle on graph paper.



  1. Find the midpoint and slope ofABand use them to draw the perpendicular bisector ofAB. You do not need to
    write the equation.

  2. Find the midpoint and slope ofBCand use them to draw the perpendicular bisector ofBC. You do not need to
    write the equation.

  3. Find the midpoint and slope ofACand use them to draw the perpendicular bisector ofAC. You do not need to
    write the equation.

  4. Are the three lines concurrent? What are the coordinates of their point of intersection (what is the circumcenter
    of the triangle)?

  5. Use your compass to draw the circumscribed circle about the triangle with your point found in question 11 as
    the center of your circle.

  6. Fill in the blanks: There is exactly _____ circle which contains any __ points.

  7. Fill in the blanks of the proof of the Perpendicular Bisector Theorem.


Given:


←→


CDis the perpendicular bisector ofAB

Prove:AC∼=CB


TABLE5.2:


Statement Reason
1.
2.Dis the midpoint ofAB


  1. Definition of a midpoint
    4.^6 CDAand^6 CDBare right angles
    5.^6 CDA∼=^6 CDB

  2. Reflexive PoC

  3. 4 CDA∼= 4 CDB
    8.AC∼=CB

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