The vertices of the triangle are A(5, 4, 6), B(1, -1, 3) and C(4, 3, 2). The internal bisector of angle A meets BC at D. Find the coordinates of D and the length AD.

Asked by Sakshi | 1 year ago |  36

##### Solution :-

Given:

The vertices of the triangle are A (5, 4, 6), B (1, -1, 3) and C (4, 3, 2).

By using the formulas let us find the coordinates of D and the length of AD

The distance between any two points (a, b, c) and (m, n, o) is given by,

The distance between A (5, 4, 6), B (1, -1, 3) is AB

$$\sqrt{(5-2)^2+4-(-1)^2+(6-3)^2}$$

$$\sqrt{16+25+9}=5\sqrt{2}$$

The distance between A (5, 4, 6), C (4, 3, 2) is AC

$$\sqrt{(5-4)^2+(4-3)^2+(6-2)^2}$$

$$\sqrt{1+1+16}=3\sqrt{2}$$

AB : AC = 5:3

BD: DC = 5:3

So, m = 5 and n = 3

B(1, -1, 3) and C(4, 3, 2)

Coordinates of D using section formula:

The coordinates of D are $$( \dfrac{23}{8},\dfrac{3}{2},\dfrac{19}{8})$$

Answered by Aaryan | 1 year ago

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