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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Aicharya · Accepted Answer

The triangle formed by the lines $y=\sqrt{3} x, y=-\sqrt{3} x$ and $y=3$ is shown in the figure.

We can observe that the triangle $A B C$ is an isosceles triangle. Therefore, the incentre lie on the median to the base. Since $D$ is mid-point of $B C$. So, $O D$ is median to the base $B C$. Thus, incentre of $\triangle A B C$ lie on $Y$-axis.

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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