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Q.
If the equation of the base of an equilateral triangle is $x+y=2$ and the vertex is $(2,-1)$, then the length of the side of the triangle is
Straight Lines
Solution:
Let $A B C$ be an equilateral triangle. Draw $A D \perp B C$.
Length of $A D=$ Perpendicular distance from $A(2,-1)$ to the line $x+y-2=0$
$=\left|\frac{2-1-2}{\sqrt{1+1}}\right|=\frac{1}{\sqrt{2}}$
Since, $\triangle A B C$ is an equilateral triangle, so $\angle A B C=60^{\circ}$.
In $\triangle A B D$,
$\Rightarrow \sin 60^{\circ}=\frac{A D}{A B}$
$ \Rightarrow \frac{\sqrt{3}}{2}=\frac{\frac{1}{\sqrt{2}}}{A B} $
$\Rightarrow A B=\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{2}} A B$
$\Rightarrow A B =\sqrt{\frac{2}{3}} $ unit
Since, $\triangle A B C$ is an equilateral triangle, so all sides are equal length.
$\therefore $ Length of $ A B=$ Length of $ B C=$ Length of $C A=\sqrt{\frac{2}{3}}$
