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 $B C$ at the point $D$. Find the coordinates of $D$.

Home » Let A and C be the vertices of a triangle. The internal bisector of the angle meets at the point . Find the coordinates of .

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 $B C$ at the point $D$ . Find the coordinates of $D$ .

CBSE | Class 11 | Introduction to Three Dimensional Geometry | Maths | NCERT Exemplar

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 $B C$ at the point $D$ . Find the coordinates of $D$ .

Solution:

It is given $A(2,2,-3), B(5,6,9)$ and $C(2,7,9)$ are the vertices of a triangle. And it is also given that the internal bisector of the angle A meets BC at the point D.
$\mathrm{AB}=\sqrt{(5-2)^{2}+(6-2)^{2}+(9-(-3))^{2}}=\sqrt{9+16+144}=\sqrt{169}=13$
So 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 therefore the internal bisector of the angle $\mathrm{A}$ meets $\mathrm{BC}$ 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 $\left(\frac{7}{2}, \frac{13}{2}, 9\right)$