A vertex of an equilateral triangle is (2, 3) and the opposite side is x + y = 2. Find the equations of the other sides.

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Solution

Dear student
$L e t A (2, 3) b e o n e v e r t e x a n d x + y = 2 b e t h e o p p o s i t e s i d e o f t h e g i v e n e q u i l a t e r a l t r i a n g l e . H e n c e t h e o t h e r t w o s i d e s p a s s t h r o u g h t h e p o i n t A (2, 3) a n d m a k e a n a n g l e 60 . S l o p e o f t h e l i n e x + y = 2 i s 1 . m = - 1 ∴ E q u a t i o n o f t h e o t h e r t w o s i d e s a r e y - 3 = (\frac{- 1 - \tan 60 °}{1 - \tan 60 °}) (x - 2) a n d y - 3 = (\frac{- 1 - \tan 60 °}{1 + \tan 60 °}) (x - 2) B u t \tan 60 ° = \sqrt{3} ∴ y - 3 = \frac{- 1 - \sqrt{3}}{1 - \sqrt{3}} (x - 2) \Rightarrow y - 3 = \frac{- (1 + \sqrt{3})}{1 - \sqrt{3}} (x - 2) \Rightarrow y - 3 = (2 + \sqrt{3}) (x - 2) i s t h e e q u a t i o n o f o n e s i d e . y - 3 = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} (x - 2) \Rightarrow (y - 3) (2 + \sqrt{3}) = (x - 2) i s t h e e q u a t i o n o f t h e o t h e r s i d e . \Rightarrow y - 3 = (2 - \sqrt{3}) (x - 2)$
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