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

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

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$(2, 2)$

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(B) The given line is $3x + 4y = 24$. To find the $x-$intercept $(A)$,set $y = 0$: $3x = 24 \implies x = 8$. So,$A = (8, 0)$. To find the $y-$intercept $(B)$,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 $AB = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = 10$. The incentre $I(x, y)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,$(x_3, y_3)$ and opposite side lengths $a, b, c$ is given by $\left( \frac{ax_1 + bx_2 + cx_3}{a + b + c}, \frac{ay_1 + by_2 + cy_3}{a + b + c} \right)$. Here,$x_1=0, y_1=0$ (opposite side $a=AB=10$),$x_2=8, y_2=0$ (opposite side $b=OB=6$),$x_3=0, y_3=6$ (opposite side $c=OA=8$). $I = \left( \frac{10(0) + 6(8) + 8(0)}{10 + 6 + 8}, \frac{10(0) + 6(0) + 8(6)}{10 + 6 + 8} \right) = \left( \frac{48}{24}, \frac{48}{24} \right) = (2, 2)$.

If the line $3x + 4y - 24 = 0$ intersects the $x-$axis at the point $A$ and the $y-$axis at the point $B$,then the incentre of the triangle $OAB$,where $O$ is the origin,is

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