x+8y-22=0, 5x+2y-34=0, and 2x-3y+13=0 are the | vedclass

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(B) Given the three sides of a triangle are:$x+8y-22=0$ $(i)$$5x+2y-34=0$ $(ii)$$2x-3y+13=0$ $(iii)$Solving $(i)$ and $(ii)$:$x=6, y=2$. Vertex $A = (

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$19$ sq units

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(B) Given the three sides of a triangle are:$x+8y-22=0$ $(i)$$5x+2y-34=0$ $(ii)$$2x-3y+13=0$ $(iii)$Solving $(i)$ and $(ii)$:$x=6, y=2$. Vertex $A = (6, 2)$.Solving $(ii)$ and $(iii)$:$x=4, y=7$. Vertex $B = (4, 7)$.Solving $(i)$ and $(iii)$:$x=-2, y=3$. Vertex $C = (-2, 3)$.The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by:$	ext{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$$	ext{Area} = \frac{1}{2} |6(7-3) + 4(3-2) + (-2)(2-7)|$$	ext{Area} = \frac{1}{2} |6(4) + 4(1) - 2(-5)|$$	ext{Area} = \frac{1}{2} |24 + 4 + 10|$$	ext{Area} = \frac{1}{2} |38| = 19$ sq units.

$x+8y-22=0$, $5x+2y-34=0$, and $2x-3y+13=0$ are the three sides of a triangle. The area of the triangle is

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