The incenter of the triangle formed by the po | vedclass

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Question

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

Vedclass · Accepted Answer

$(1,1,0)$

Vedclass · Answer

(A) Let the vertices of the triangle be $A(0,0,0), B(3,0,0)$,and $C(0,4,0)$.First,we calculate the lengths of the sides opposite to the vertices $A, B$,and $C$:$a = BC = \sqrt{(0-3)^2 + (4-0)^2 + (0-0)^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$$b = AC = \sqrt{(0-0)^2 + (4-0)^2 + (0-0)^2} = \sqrt{4^2} = 4$$c = AB = \sqrt{(3-0)^2 + (0-0)^2 + (0-0)^2} = \sqrt{3^2} = 3$The coordinates of the incenter $(x, y, z)$ are given by the formula:$x = \frac{ax_1 + bx_2 + cx_3}{a+b+c}, y = \frac{ay_1 + by_2 + cy_3}{a+b+c}, z = \frac{az_1 + bz_2 + cz_3}{a+b+c}$Substituting the values $A(x_1, y_1, z_1) = (0,0,0), B(x_2, y_2, z_2) = (3,0,0), C(x_3, y_3, z_3) = (0,4,0)$ and side lengths $a=5, b=4, c=3$:$x = \frac{5(0) + 4(3) + 3(0)}{5+4+3} = \frac{12}{12} = 1$$y = \frac{5(0) + 4(0) + 3(4)}{5+4+3} = \frac{12}{12} = 1$$z = \frac{5(0) + 4(0) + 3(0)}{5+4+3} = \frac{0}{12} = 0$Thus,the incenter is $(1, 1, 0)$.

The incenter of the triangle formed by the points $(0,0,0), (3,0,0)$ and $(0,4,0)$ is

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