Question

Prove that the lines $y=\sqrt{3}x+1,y=4andy=-\sqrt{3}x+2$ form an equilateral triangle.

Answer 1

$\text{Let the line be}$

$y=\sqrt{3}x+1...\left(1\right)$

$y=4...\left(2\right)$

$y=-\sqrt{3}x+2...\left(3\right)$

$\text{Solve (1) and (2)}$

$4=\sqrt{3}x+1$

$x=\frac{4-1}{\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}$

$\therefore \text{Point A is}(\sqrt{3},4)$

$\text{Solve (2) and (3)}$

$4=-\sqrt{3}x+2$

$\sqrt{3}x=-2$

$x=\frac{-2}{\sqrt{3}}$

$=\frac{-2\sqrt{3}}{3}$

$\therefore \text{Point B is}(\frac{-2\sqrt{3}}{3},4)$

$\text{Solve (1) and (3)}$

$\sqrt{3}x+1=-\sqrt{3}x+2$

$2\sqrt{3}x=1$

$x=\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}$

$y=\sqrt{3}\left(\frac{\sqrt{3}}{6}\right)+1$

$=\frac{3}{2}$

$\therefore \text{Point C is}(\frac{\sqrt{3}}{6},\frac{3}{2})$

$\text{length}l=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$let{l}_1=\text{length of AB}$

$=\sqrt{{(\frac{-2\sqrt{3}}{3}-\sqrt{3})}^2+(4-4{)}^2}$

$=\sqrt{{\left(\frac{5\sqrt{3}}{3}\right)}^2}$

$=\frac{5\sqrt{3}}{3}\text{units}$

$\text{let length}{l}_2=\text{length of BC}$

$=\sqrt{{(\frac{\sqrt{3}}{6}+\frac{2\sqrt{3}}{3})}^2+{(\frac{3}{2}-4)}^2}$

$=\sqrt{{\left(\frac{5\sqrt{3}}{6}\right)}^2+{\left(\frac{-5}{2}\right)}^2}$

$=\sqrt{\frac{75}{36}+\frac{25}{4}}$

$=\sqrt{\frac{75}{36}+\frac{255}{36}}$

$=\sqrt{\frac{300}{36}}$

$=\frac{10}{6}\sqrt{3}$

$=\frac{5}{3}\sqrt{3}.$

$\text{let length}{l}_3=\text{length of AC}$

$=\sqrt{{(\frac{\sqrt{3}}{6}-\sqrt{3})}^2+{(\frac{3}{2}-4)}^2}$

$=\sqrt{{\left(\frac{5\sqrt{3}}{6}\right)}^2+{\left(\frac{-5}{2}\right)}^2}$

$=\sqrt{\frac{75}{36}+\frac{25}{4}}$

$=\sqrt{\frac{75}{36}+\frac{225}{36}}$

$=\sqrt{\frac{300}{36}}$

$=\frac{10}{6}\sqrt{3}$

$=\frac{5}{3}\sqrt{3}$

$\Rightarrow l_1=l_2=l_3$

$\text{Hence, triangle is equilateral.}$