The incentre of the triangle formed by the li | vedclass

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(B) The lines $x = 0$ (y-axis),$y = 0$ (x-axis),and $3x + 4y = 12$ form a right-angled triangle with vertices at $A(0, 0)$,$B(0, 3)$,and $C(4, 0)$.Let

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$(1, 1)$

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(B) The lines $x = 0$ (y-axis),$y = 0$ (x-axis),and $3x + 4y = 12$ form a right-angled triangle with vertices at $A(0, 0)$,$B(0, 3)$,and $C(4, 0)$.Let the vertices be $A(x_1, y_1) = (0, 0)$,$B(x_2, y_2) = (0, 3)$,and $C(x_3, y_3) = (4, 0)$.The lengths of the sides opposite to these vertices are:$a = BC = \sqrt{(4-0)^2 + (0-3)^2} = \sqrt{16 + 9} = 5$$b = AC = \sqrt{(4-0)^2 + (0-0)^2} = 4$$c = AB = \sqrt{(0-0)^2 + (3-0)^2} = 3$The coordinates of the incentre $(I_x, I_y)$ are given by:$I_x = \frac{ax_1 + bx_2 + cx_3}{a + b + c} = \frac{5(0) + 4(0) + 3(4)}{5 + 4 + 3} = \frac{12}{12} = 1$$I_y = \frac{ay_1 + by_2 + cy_3}{a + b + c} = \frac{5(0) + 4(3) + 3(0)}{5 + 4 + 3} = \frac{12}{12} = 1$Thus,the incentre is $(1, 1)$.

The incentre of the triangle formed by the lines $x = 0$,$y = 0$,and $3x + 4y = 12$ is

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