The incentre of the triangle formed by the st | vedclass

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(A) The vertices of the triangle are the intersection points of the lines:$1$. $y=\sqrt{3}x$ and $y=-\sqrt{3}x$ intersect at $A(0, 0)$.$2$. $y=\sqrt{3

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$(0, 2)$

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(A) The vertices of the triangle are the intersection points of the lines:$1$. $y=\sqrt{3}x$ and $y=-\sqrt{3}x$ intersect at $A(0, 0)$.$2$. $y=\sqrt{3}x$ and $y=3$ intersect at $B(\sqrt{3}, 3)$.$3$. $y=-\sqrt{3}x$ and $y=3$ intersect at $C(-\sqrt{3}, 3)$.The lengths of the sides are:$c = AB = \sqrt{(\sqrt{3}-0)^2 + (3-0)^2} = \sqrt{3+9} = \sqrt{12} = 2\sqrt{3}$.$b = AC = \sqrt{(-\sqrt{3}-0)^2 + (3-0)^2} = \sqrt{3+9} = 2\sqrt{3}$.$a = BC = \sqrt{(\sqrt{3}-(-\sqrt{3}))^2 + (3-3)^2} = \sqrt{(2\sqrt{3})^2} = 2\sqrt{3}$.Since $a=b=c$,the triangle is equilateral.The incentre $(I)$ of an equilateral triangle is the same as its centroid $(G)$.$I = G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{0+\sqrt{3}-\sqrt{3}}{3}, \frac{0+3+3}{3}\right) = \left(0, \frac{6}{3}\right) = (0, 2)$.

The incentre of the triangle formed by the straight lines $y=\sqrt{3}x$,$y=-\sqrt{3}x$ and $y=3$ is

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