Let A (2, 2, – 3), B (5, 6, 9) and C (2, 7, 9) be the vertices of a triangle.

Question:

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.

Solution:

Given $A(2,2,-3), B(5,6,9)$ and $C(2,7,9)$ are the vertices of a triangle. And also given that the internal bisector of the angle $A$ meets $B C$ at the point D.

$A B=\sqrt{(5-2)^{2}+(6-2)^{2}+(9-(-3))^{2}}=\sqrt{9+16+144}=\sqrt{169}=13$

Now,

$\mathrm{AC}=\sqrt{(2-2)^{2}+(7-2)^{2}+(9-(-3))^{2}}=\sqrt{0+25+144}=\sqrt{169}=13$

$\Rightarrow \mathrm{ABC}$ is an isosceles triangle and thus the internal bisector of the angle $\mathrm{A}$ meets BC at its midpoint.

 $\Rightarrow \mathrm{D}\left(\frac{5+2}{2}, \frac{6+7}{2}, \frac{9+9}{2}\right)$

$\therefore$ The coordinates of $D$ is $\left(\frac{7}{2}, \frac{13}{2}, 9\right)$

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