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(A) The given lines are $x=0$ (the $y$-axis),$y=0$ (the $x$-axis),and $3x+4y=12$.To find the intercepts of the line $3x+4y=12$,we rewrite it in the in

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$6$

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(A) The given lines are $x=0$ (the $y$-axis),$y=0$ (the $x$-axis),and $3x+4y=12$.To find the intercepts of the line $3x+4y=12$,we rewrite it in the intercept form $\frac{x}{a} + \frac{y}{b} = 1$.Dividing the equation by $12$,we get $\frac{3x}{12} + \frac{4y}{12} = \frac{12}{12}$,which simplifies to $\frac{x}{4} + \frac{y}{3} = 1$.This means the line intersects the $x$-axis at point $B(4, 0)$ and the $y$-axis at point $A(0, 3)$.The triangle formed by these lines is a right-angled triangle with vertices at $O(0, 0)$,$A(0, 3)$,and $B(4, 0)$.The base of the triangle is $OB = 4$ units and the height is $OA = 3$ units.The area of the triangle is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4 	imes 3 = 6$ square units.

The area (in square units) of the triangle formed by the lines $x=0, y=0$ and $3x+4y=12$ is

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