Question on Quadratic Equations.11 Base of an equilateral triangle lies along the line $$9x+40y-50=0$$ and its vertex opposite to the base lies on the line $$9x+40y+32=0$$ Find the length of the side of the triangle and also find its area.
 A: We know if the length of a side of the triangle $=a$ and the height $=d$
$d=\dfrac{\sqrt3}2a$
and here $d=\dfrac{|32-(-50)|}{\sqrt{9^2+4^2}}$
$d$ can also be calculated as follows:
Find any arbitrary point $P(h,k)$ on any one of the straight line 
and find the perpendicular distance of $P$ from the other  straight line.
As $\left(0,\dfrac54\right)$ is on $9x+40y-50=0$
$d$ will be the perpendicular distance of  $\left(0,\dfrac54\right)$ from $9x+40y+32=0$
A: Hint Note that these lines are parallel and so you can determine the height of the triangle by computing the distance between these lines (= length of an orthogonal segment from one line to the other). Once you have the height, you can use Pythagoras Theorem to find the length of one side and finally determine its area.
A: The distance between the parallel lines: $9x+40y-50=0$ & $9x+40y+32=0$  is $$\frac{|(-50)-(32)|}{\sqrt{(9)^2+(40)^2}}=\frac{82}{\sqrt{1681}}=\frac{82}{41}=2$$ The above distance between the parallel sides will be equal to the vertical height of equilateral triangle. 
Let the side of the equilateral triangle be $a$ then its vertical height $$=\frac{\sqrt{3}}{2}a=2$$ $$\implies \color{blue}{\text{side},\ a=\frac{4}{\sqrt{3}}}$$ Hence the area of equilateral triangle $$=\frac{\sqrt3}{4}(a^2)$$ $$=\frac{\sqrt3}{4}\left(\frac{4}{\sqrt 3}\right)^2$$ $$=\color{blue}{\frac{4}{\sqrt3}}$$
