Show that the lines x² - 4xy + y² = 0 and x + y = √6 form an equilateral triangle. Find its area and perimeter.
Answers
Given : lines x² - 4xy + y² = 0 and x + y = √6 form an equilateral triangle.
To find : area and perimeter of triangle
Solution:
x² - 4xy + y² = 0
=> (x - y)² -2xy = 0
=> (x - y)² = 2xy
x = 0 , y = 0 Satisfy this
x + y = √6 & x² - 4xy + y² = 0
=> x² - 4x(√6 - x) + (√6 - x)² = 0
=> x² - 4x√6 + 4x² + 6 + x² - 2x√6 = 0
=> 6x² - 6x√6 + 6 = 0
=> x² - x√6 + 1 = 0
=> x = (√6 ± √2)/2
x = (√6 + √2)/2 , (√6 - √2)/2
using x + y = √6
y = (√6 - √2)/2 , (√6 + √2)/2
(0 , 0) , ( (√6 + √2)/2 , (√6 - √2)/2 ) & ( (√6- √2)/2 , (√6 + √2)/2 )
are three vertex of triangle
(0 , 0) , ( (√6 + √2)/2 , (√6 - √2)/2 ) = √ (6 + 2 + 2√12)/4 + (6 + 2 - 2√12)/4 = √ 4 = 2
(0 , 0) , ( (√6- √2)/2 , (√6 + √2)/2 ) = √ (6 + 2 - 2√12)/4 + (6 + 2 +2√12)/4 = √ 4 = 2
( (√6 + √2)/2 , (√6 - √2)/2 ) & ( (√6- √2)/2 , (√6 + √2)/2 )
= √ (-√2)² + (√2)² = √4 = 2
Hence length of all Sides = 2
Perimeter = 2 + 2 + 2 = 6 unit
Area = (√3 / 4)(2)² = √3 sq unit
Perimeter = 6 unit & area= √3 sq unit
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