If the line 3x + 4y - 24 = 0 intersects X and | vedclass

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(B) The given line is $3x + 4y = 24$.To find the intersection with the $X$-axis,set $y = 0$: $3x = 24 \implies x = 8$. So,$A = (8, 0)$.To find the int

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$(2, 2)$

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(B) The given line is $3x + 4y = 24$.To find the intersection with the $X$-axis,set $y = 0$: $3x = 24 \implies x = 8$. So,$A = (8, 0)$.To find the intersection with the $Y$-axis,set $x = 0$: $4y = 24 \implies y = 6$. So,$B = (0, 6)$.The vertices of the triangle $OAB$ are $O(0, 0)$,$A(8, 0)$,and $B(0, 6)$.The lengths of the sides are $OA = 8$,$OB = 6$,and $AB = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$.The incenter $(I_x, I_y)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ and opposite side lengths $a, b, c$ is given by $(\frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c})$.Here,$x_1=0, y_1=0$ (opposite side $a=10$); $x_2=8, y_2=0$ (opposite side $b=6$); $x_3=0, y_3=6$ (opposite side $c=8$).$I_x = \frac{10(0) + 6(8) + 8(0)}{10 + 6 + 8} = \frac{48}{24} = 2$.$I_y = \frac{10(0) + 6(0) + 8(6)}{10 + 6 + 8} = \frac{48}{24} = 2$.Thus,the incenter is $(2, 2)$.

If the line $3x + 4y - 24 = 0$ intersects $X$ and $Y$ axes at points $A$ and $B$ respectively,then the incenter of the triangle $OAB$,where $O$ is the origin,is:

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