If A(0,0,0), B(3,4,0), C(0,12,5) are the vert | vedclass

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(C) Let the vertices be $A(0,0,0)$,$B(3,4,0)$,and $C(0,12,5)$.First,we calculate the lengths of the sides opposite to the vertices $A, B, C$ denoted b

Vedclass · Accepted Answer

$\frac{39}{18+7 \sqrt{2}}$

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(C) Let the vertices be $A(0,0,0)$,$B(3,4,0)$,and $C(0,12,5)$.First,we calculate the lengths of the sides opposite to the vertices $A, B, C$ denoted by $a, b, c$ respectively.$a = BC = \sqrt{(0-3)^2 + (12-4)^2 + (5-0)^2} = \sqrt{(-3)^2 + 8^2 + 5^2} = \sqrt{9 + 64 + 25} = \sqrt{98} = 7\sqrt{2}$.$b = AC = \sqrt{(0-0)^2 + (12-0)^2 + (5-0)^2} = \sqrt{0^2 + 12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$.$c = AB = \sqrt{(3-0)^2 + (4-0)^2 + (0-0)^2} = \sqrt{3^2 + 4^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.The $x$-coordinate of the incentre $I(x, y, z)$ is given by the formula $x = \frac{ax_A + bx_B + cx_C}{a+b+c}$.Substituting the values: $x = \frac{(7\sqrt{2})(0) + (13)(3) + (5)(0)}{7\sqrt{2} + 13 + 5} = \frac{0 + 39 + 0}{18 + 7\sqrt{2}} = \frac{39}{18 + 7\sqrt{2}}$.

If $A(0,0,0), B(3,4,0), C(0,12,5)$ are the vertices of a triangle $ABC$,then the $x$-coordinate of its incentre is

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