Question

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

Answer 1

The given equations are as follows:

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

y = 4 ... (2)

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

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2):
$x=\sqrt{3}$, y = 4

Thus, AB and BC intersect at $B\left(\sqrt{3},4\right)$.

Solving (1) and (3):
$x=\frac{1}{2\sqrt{3}},y=\frac{3}{2}$

Thus, AB and CA intersect at A $\left(\frac{1}{2\sqrt{3}},\frac{3}{2}\right)$.

Similarly, solving (2) and (3):
$x=-\frac{2}{\sqrt{3}},y=4$

Thus, BC and AC intersect at $C\left(-\frac{2}{\sqrt{3}},4\right)$.

Now, we have:

$$\begin{gathered} AB=\sqrt{{\left(\frac{1}{2\sqrt{3}}-\sqrt{3}\right)}^2+{\left(\frac{3}{2}-4\right)}^2}=\frac{5}{\sqrt{3}} \\ BC=\sqrt{{\left(\sqrt{3}+\frac{2}{\sqrt{3}}\right)}^2+{\left(4-4\right)}^2}=\frac{5}{\sqrt{3}} \\ AC=\sqrt{{\left(\frac{1}{2\sqrt{3}}+\frac{2}{\sqrt{3}}\right)}^2+{\left(\frac{3}{2}-4\right)}^2}=\frac{5}{\sqrt{3}} \end{gathered}$$

Hence, the given lines form an equilateral triangle.