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Question
Prove that the lines \[y = \sqrt{3}x + 1, y = 4 \text { and } y = - \sqrt{3}x + 2\] form an equilateral triangle.
Solution
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):
\[AB = \sqrt{\left( \frac{1}{2\sqrt{3}} - \sqrt{3} \right)^2 + \left( \frac{3}{2} - 4 \right)^2} = \frac{5}{\sqrt{3}}\]
\[BC = \sqrt{\left( \sqrt{3} + \frac{2}{\sqrt{3}} \right)^2 + \left( 4 - 4 \right)^2} = \frac{5}{\sqrt{3}}\]
\[AC = \sqrt{\left( \frac{1}{2\sqrt{3}} + \frac{2}{\sqrt{3}} \right)^2 + \left( \frac{3}{2} - 4 \right)^2} = \frac{5}{\sqrt{3}}\]
Hence, the given lines form an equilateral triangle.
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