The incentre of the triangle whose vertices a | vedclass

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(B) Let the vertices be $P(0,3,0)$,$Q(0,0,4)$,and $R(0,3,4)$.The side lengths are:$PQ = \sqrt{(0-0)^2 + (0-3)^2 + (4-0)^2} = \sqrt{0+9+16} = 5$$QR = \

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$(0,2,3)$

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(B) Let the vertices be $P(0,3,0)$,$Q(0,0,4)$,and $R(0,3,4)$.The side lengths are:$PQ = \sqrt{(0-0)^2 + (0-3)^2 + (4-0)^2} = \sqrt{0+9+16} = 5$$QR = \sqrt{(0-0)^2 + (3-0)^2 + (4-4)^2} = \sqrt{0+9+0} = 3$$PR = \sqrt{(0-0)^2 + (3-3)^2 + (4-0)^2} = \sqrt{0+0+16} = 4$The incentre $I(x,y,z)$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ and opposite side lengths $a, b, c$ is given by:$I = \left( \frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c}, \frac{az_1 + bz_2 + cz_3}{a+b+c} \right)$Here,$a=QR=3$,$b=PR=4$,$c=PQ=5$.$I = \left( \frac{3(0)+4(0)+5(0)}{3+4+5}, \frac{3(3)+4(0)+5(3)}{3+4+5}, \frac{3(0)+4(4)+5(4)}{3+4+5} \right)$$I = \left( \frac{0}{12}, \frac{9+0+15}{12}, \frac{0+16+20}{12} \right)$$I = \left( 0, \frac{24}{12}, \frac{36}{12} \right) = (0, 2, 3)$

The incentre of the triangle whose vertices are $P(0,3,0)$,$Q(0,0,4)$,and $R(0,3,4)$ is

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