If the equation of the base of an equilateral | vedclass

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(C) The length of the altitude $h$ from the vertex $(-1, -1)$ to the line $x+y-6=0$ is given by the perpendicular distance formula: $h = \frac{|(-1) +

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$\frac{32}{\sqrt{3}}$

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(C) The length of the altitude $h$ from the vertex $(-1, -1)$ to the line $x+y-6=0$ is given by the perpendicular distance formula: $h = \frac{|(-1) + (-1) - 6|}{\sqrt{1^2 + 1^2}} = \frac{|-8|}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}$.In an equilateral triangle with side length $a$,the altitude $h = \frac{\sqrt{3}}{2}a$. Therefore,$a = \frac{2h}{\sqrt{3}} = \frac{2(4\sqrt{2})}{\sqrt{3}} = \frac{8\sqrt{2}}{\sqrt{3}}$.The area of the equilateral triangle is $A = \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4} \left(\frac{8\sqrt{2}}{\sqrt{3}}ight)^2 = \frac{\sqrt{3}}{4} 	imes \frac{64 	imes 2}{3} = \frac{\sqrt{3}}{4} 	imes \frac{128}{3} = \frac{32\sqrt{3}}{3} = \frac{32}{\sqrt{3}}$ sq. units.

If the equation of the base of an equilateral triangle is $x+y=6$ and the opposite vertex is the point $(-1, -1)$,then the area of the triangle is equal to $k$ sq. units. Find the value of $k$.

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