The in-centre of the triangle formed by the l | vedclass

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(A) The lines are $x=0$ (y-axis),$y=0$ (x-axis),and $3x+4y=12$. The vertices of the triangle are the intersection points of these lines: $1$. Intersec

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$(1, 1)$

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(A) The lines are $x=0$ (y-axis),$y=0$ (x-axis),and $3x+4y=12$. The vertices of the triangle are the intersection points of these lines: $1$. Intersection of $x=0$ and $y=0$ is $A(0, 0)$. $2$. Intersection of $x=0$ and $3x+4y=12$ is $B(0, 3)$. $3$. Intersection of $y=0$ and $3x+4y=12$ is $C(4, 0)$. The side lengths 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 in-centre $(I_x, I_y)$ is given by $\left(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c}\right)$. Substituting the values: $I_x = \frac{5(0) + 4(0) + 3(4)}{5+4+3} = \frac{12}{12} = 1$. $I_y = \frac{5(0) + 4(3) + 3(0)}{5+4+3} = \frac{12}{12} = 1$. Thus,the in-centre is $(1, 1)$.

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

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