Question

Determine the point of intersection of right bisectors in a triangle ∆ABC with vertices A (-3, 5), B (1, 1) and C (−7, −3). Find the distance from the point of intersection to each vertex of the triangle.

Answer 1

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Answer:

• intersection: (-3 2/3, 1/3)
• distances to vertices: (10/3)√2

Step-by-step explanation:

A) In order to write the equations of the perpendicular bisectors of the segments, we need to know the midpoint and the slope.

The midpoint is the average of the coordinates:

  midpoint(A, B) = ((-3, 5) +(1, 1))/2 = (-3+1, 5+1)/2 = (-1, 3)

  midpoint(A, C) = ((-3, 5) +(-7, -3))/2 = (-3-7, 5-3)/2 = (-5, 1)

The differences between the coordinates of the points can be helpful:

  B-A = (1, 1) -(-3, 5) = (1+3, 1-5) = (4, -4) . . . . . reduces to (∆x, ∆y) = (1, -1)

  C-A = (-7, -3) -(-3, 5) = (-7+3, -3-5) = (-4, -8) . . . . . reduces to (∆x, ∆y) = (1, 2)

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A line through point (h, k) perpendicular to one with (∆x, ∆y) can be written:

  (∆x)(x -h) +(∆y)(y -k) = 0

Then the bisectors of AB and AC will be ...

  (x +1) -(y -3) = 0

  (x +5) +2(y -1) = 0

Their intersection point can be found by solving this system.

  (x +2y +3) -(x - y + 4) = (0) -(0) . . . . . subtract the 1st equation from the 2nd

  3y -1 = 0

  y = 1/3

  x = y -4 = -3 2/3

The point of intersection of the right bisectors is (x, y) = (-3 2/3, 1/3).

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B) The intersection point is the circumcenter. It will be the same distance from any of the vertices.

That distance can be found using any of the vertices. Using point B, we find the distance to be ...

  d = √((-3 2/3 -1)^2 +(1/3 -1)^2) = √((14/3)^2 +(2/3)^2) = √(200/9)

  d = (10/3)√2

The distance to each vertex is (10/3)√2 ≈ 4.71.