The incentre of the triangle formed by the st | vedclass

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(D) The vertices of the triangle are $A(0, 0)$,$B(3, 0)$,and $C(0, 4)$.Let the side lengths be $a$,$b$,and $c$ opposite to vertices $A$,$B$,and $C$ re

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$(1, 1)$

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(D) The vertices of the triangle are $A(0, 0)$,$B(3, 0)$,and $C(0, 4)$.Let the side lengths be $a$,$b$,and $c$ opposite to vertices $A$,$B$,and $C$ respectively.$a = BC = \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.$b = AC = \sqrt{(0-0)^2 + (4-0)^2} = 4$.$c = AB = \sqrt{(3-0)^2 + (0-0)^2} = 3$.The coordinates of the incentre $I$ are given by $\left(\frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c}\right)$.Here,$(x_1, y_1) = (0, 0)$,$(x_2, y_2) = (3, 0)$,and $(x_3, y_3) = (0, 4)$.$I = \left(\frac{5(0) + 4(3) + 3(0)}{5+4+3}, \frac{5(0) + 4(0) + 3(4)}{5+4+3}\right)$.$I = \left(\frac{12}{12}, \frac{12}{12}\right) = (1, 1)$.

The incentre of the triangle formed by the straight line having $3$ as $X$-intercept and $4$ as $Y$-intercept,together with the coordinate axes,is

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