The incentre of the triangle formed by $(0, 0, 0), (3, 0, 0), (0, 3, 0)$ .

(\frac{3}{4}, \frac{3}{4}, 0)

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(1, \frac{3}{4}, 0)

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(0, 1, 1)

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(1, 1, 1)

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Solution

The correct option is D $(1, \frac{3}{4}, 0)$
using formula incenter $x = \frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}$

$y = \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c}$
$z = \frac{a z_{1} + b z_{2} + c x_{3}}{a + b + c}$
$a = 5$ $x = \frac{5 (0) + 4 (3) + 3 (0)}{5 + 4 + 3}$ ; $x = 1$

$b = 4$ $y = \frac{5 (0) + 4 (0) + 3 (3)}{5 + 4 + 3}$ ; $y = \frac{3}{4}$

$c = 3$ $z = \frac{5 (0) + 4 (0) + 3 (0)}{5 + 4 + 3}$ ; $z = 0$

incentre = $(1, \frac{3}{4}, 0)$

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The incentre of the triangle formed by (0,0,0),(3,0,0),(0,3,0).

The correct option is D (1,34,0)using formula incenter x=ax1+bx2+cx3a+b+c y=ay1+by2+cy3a+b+c z=az1+bz2+cx3a+b+c a=5 x=5(0)+4(3)+3(0)5+4+3 ; x=1 b=4 y=5(0)+4(0)+3(3)5+4+3 ; y=34 c=3 z=5(0)+4(0)+3(0)5+4+3 ; z=0 incentre =(1,34,0)

The incentre of the triangle formed by $(0, 0, 0), (3, 0, 0), (0, 3, 0)$ .