Question

Prove that the lines  \[y = \sqrt{3}x + 1, y = 4 \text { and } y = - \sqrt{3}x + 2\] 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:

\[AB = \sqrt{\left( \frac{1}{2\sqrt{3}} - \sqrt{3} \right)^2 + \left( \frac{3}{2} - 4 \right)^2} = \frac{5}{\sqrt{3}}\]

\[BC = \sqrt{\left( \sqrt{3} + \frac{2}{\sqrt{3}} \right)^2 + \left( 4 - 4 \right)^2} = \frac{5}{\sqrt{3}}\]

\[AC = \sqrt{\left( \frac{1}{2\sqrt{3}} + \frac{2}{\sqrt{3}} \right)^2 + \left( \frac{3}{2} - 4 \right)^2} = \frac{5}{\sqrt{3}}\]

Hence, the given lines form an equilateral triangle.