The area (in square units) of the triangle fo | vedclass

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(C) 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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(C) 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 $\frac{x}{a} + \frac{y}{b} = 1$:$\frac{3x}{12} + \frac{4y}{12} = \frac{12}{12} \implies \frac{x}{4} + \frac{y}{3} = 1$.This line intersects the $x$-axis at $A(4, 0)$ and the $y$-axis at $B(0, 3)$.The triangle formed by the lines $x=0, y=0$,and $3x+4y=12$ is a right-angled triangle with vertices at $O(0, 0)$,$A(4, 0)$,and $B(0, 3)$.The base of the triangle is $OA = 4$ units and the height is $OB = 3$ units.Area of $	riangle OAB = \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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