An equilateral triangle is constructed between two parallel line -3x-y-6-0 and -3x-y-9-0 with base on one and vertex on the other- Then the area of triangle is -

Given,the parallel lines are √( 3 ) x+y 6=0 ⇒ y= √( 3 ) x+6 ⟶ ( 1 ) ...

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Standard XII

Mathematics

Distance between Parallel Lines

An equilatera...

Question

An equilateral triangle is constructed between two parallel line $\sqrt{3} x + y - 6 = 0$ and $\sqrt{3} x + y + 9 = 0$ with base on one and vertex on the other. Then the area of triangle is ?

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Solution

$G i v e n, t h e p a r a l l e l l i n e s a r e - \sqrt{3} x + y - 6 = 0 \Rightarrow y = - \sqrt{3} x + 6 ⟶ (1) a n d \sqrt{3} x + y + 9 = 0 \Rightarrow y = - \sqrt{3} x - 9 ⟶ (2) D i s t a n c e b e t w e e n t w o p a r a l l e l l i n e s (1) a n d (2) i s = \frac{| c_{1} - c_{2} |}{\sqrt{1 + m^{2}}} = \frac{| 6 - (- 9) |}{\sqrt{1 + {(- \sqrt{3})}^{2}}} = \frac{15}{\sqrt{4}} = \frac{15}{2} = 7.5 F r o m t h e a b o v e f i g u r e, i t h a s b e e n s h o w n t h a t t h e d i s t a n c e b e t w e e n t h e t w o p a r a l l e l l i n e s i s e q u a l t o o n e s i d e o f a n e q u i l a t e r a l t r i a n g l e f o r m b y t w o p a r a l l e l l i n e s w i t h b a s e o n o n e a n d v e r t e x o n t h e o t h e r . ∴ T h e m e a s u r e o f s i d e s o f t h e e q u i l a t e r a l t r i a n g l e a = b = c = 7.5 u n i t s . N o w S = \frac{a + b + c}{2} = \frac{7.5 + 7.5 + 7.5}{2} = 11.25 A r e a o f t h e t r i a n g l e = \sqrt{S (S - a) (S - b) (S - c)} = \sqrt{11.25 (11.25 - 7.5) (11.25 - 7.5) (11.25 - 7.5)} = \sqrt{11.25 (3.75) (3.75) (3.75)} = \sqrt{593.261} = 24.356 s q . u n i t s / .$

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An equilateral triangle is constructed between two parallel line √3x+y−6=0 and √3x+y+9=0 with base on one and vertex on the other. Then the area of triangle is ?

An equilateral triangle is constructed between two parallel line $\sqrt{3} x + y - 6 = 0$ and $\sqrt{3} x + y + 9 = 0$ with base on one and vertex on the other. Then the area of triangle is ?