Question

let a=2,2,-3 b=5,6,9 c=2,7,9 be the vertices of triangle . the internal bisector angle A meets BC at D. find the coordinate of D

Answer 1

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Answer 2

The coordinate of D are $(\frac{7}{2}, \frac{13}{2}, 9 )$.

Step-by-step explanation:

Given:

Here, A = ( 2,2,-3 ), B = ( 5,6,9 ), C = ( 2,7,9 )

Using distance formula, we get

$AB = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

     = $\sqrt{(5-2)^2+(6-2)^2+(9-(-3)-z_1)^2}$

    = $\sqrt{(3)^2+(4)^2+(9+3)^2}$

     = $\sqrt{(3)^2+(4)^2+(12)^2}$

     = $\sqrt{9+16+144}$

     = $\sqrt{169}$

AB  = 13 units

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

     = $\sqrt{(0)^2+(5)^2+(12)^2}$

     = $\sqrt{25+144}$

     = $\sqrt{169}$

AC = 13 units

Now AD bisects ∠A , then

$\frac{AB}{AC}=\frac{BD}{DC}$      [Angle bisector theorem]

Now, AB = AC,

so, BD = DC

Therefore, BD = DC,  

⇒ D is the mid point of BC.

According to mid point formula, coordinates of D are

$D (\frac{5+3}{2}, \frac{7+6}{2}, \frac{9+9}{2} )$

$D (\frac{7}{2}, \frac{13}{2}, \frac{18}{2} )$

$D (\frac{7}{2}, \frac{13}{2}, 9 )$

Hence the coordinate of D are $(\frac{7}{2}, \frac{13}{2}, 9 )$.