The equation of base of an equilateral triangle is $x+y=2$ and the vertex is $(2,-1)$, then find the length of side of the triangle.
My Attempt :
Equation of one of the sides can be obtained as :
$$y-y_1=m(x-x_1)$$ $$y+1=\tan (60) (x-2)$$ $$y+1=\sqrt {3} (x-2)$$ $$\sqrt {3}x - y -3=0$$
Let $a$ be the length of each side of the equilateral triangle.
Given line is $x+y-2=0$
Let $d$ be the perpendicular distance from $(2,-1)$ onto the line $x+y-2=0$ $$d=\frac{1}{\sqrt{2}}$$
So, the perpendicular height of the triangle is $\dfrac{1}{\sqrt{2}}$
Now consider the one of the half side of the equilateral triangle.
According to pythagoras theorem we get, $$a^2=\frac{a^2}{4}+\left(\frac{1}{\sqrt{2}}\right)^2$$ $$\frac{3a^2}{4}=\frac12$$ $$a=\sqrt{\frac 23}$$ Therefore, the length of the side of the triangle is $\sqrt{\dfrac{2}{3}}$