If A(2, 9),B(-2, 1),and C(6, 3) are vertices | vedclass

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(D) The area of a triangle is given by the formula: $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Here,the base is the length of

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$\frac{28}{\sqrt{17}}$

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(D) The area of a triangle is given by the formula: $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Here,the base is the length of side $\overline{BC}$ and the height is the length of the altitude from $A$ to $\overline{BC}$.First,calculate the length of $\overline{BC}$ using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.For $B(-2, 1)$ and $C(6, 3)$,$BC = \sqrt{(6 - (-2))^2 + (3 - 1)^2} = \sqrt{8^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17}$.Given the area of $\Delta ABC = 28$,we have: $28 = \frac{1}{2} 	imes (2\sqrt{17}) 	imes h$.$28 = \sqrt{17} 	imes h$.Therefore,the length of the altitude $h = \frac{28}{\sqrt{17}}$.

If $A(2, 9)$,$B(-2, 1)$,and $C(6, 3)$ are vertices of a triangle and the area of $\Delta ABC$ is $28$,find the length of the altitude drawn from $A$ to the side $\overline{BC}$.

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