The area (in sq. units) bounded by x=4,y=-4 a | vedclass

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(B) The region is bounded by the lines $x=4$,$y=-4$,and $y=x$.To find the vertices of the triangle,we find the intersection points of these lines:$1$.

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) The region is bounded by the lines $x=4$,$y=-4$,and $y=x$.To find the vertices of the triangle,we find the intersection points of these lines:$1$. Intersection of $x=4$ and $y=x$ is $B(4, 4)$.$2$. Intersection of $y=-4$ and $y=x$ is $A(-4, -4)$.$3$. Intersection of $x=4$ and $y=-4$ is $C(4, -4)$.The region is a right-angled triangle $ABC$ with vertices $A(-4, -4)$,$B(4, 4)$,and $C(4, -4)$.The length of the base $AC$ is the distance between $(-4, -4)$ and $(4, -4)$,which is $|4 - (-4)| = 8$ units.The length of the height $BC$ is the distance between $(4, -4)$ and $(4, 4)$,which is $|4 - (-4)| = 8$ units.The area of the triangle is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Area $= \frac{1}{2} 	imes 8 	imes 8 = 32$ sq. units.

The area (in sq. units) bounded by $x=4$,$y=-4$ and $y=x$ is

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