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  • Let A (2, 2, - 3), B (5, 6, 9) and C (2, 7, 9) be the vertices of a triangle. The internal bisector of the angle A meets BC at the point D. Find the coordinates of D.

Let A (2, 2, – 3), B (5, 6, 9) and C (2, 7, 9) be the vertices of a triangle. The internal bisector of the angle A meets BC at the point D. Find the coordinates of D.

Answers (1)

Let us assume that the coordinates of D are (x,y,z).

Now,

AB = \sqrt{(5-2)^{2}+(6-2)^{2}+(9+3)^{2} }

                  = \sqrt{9+16+144 }  = 13

AC =\sqrt{(2-2)^{2}+(7-2)^{2}+(9+3)^{2} }

                  = \sqrt{0+25+144 } = 13

Thus, AB = AC.

Thus, ABC is an isosceles triangle.

Now, D is the mid-point of BC or Angle bisector AD bisects BD,

Thus, D = (5+2/2, 6+7/2, 9+9/2)

                  = (7/2, 13/2,9)

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