Prove that the points A(2, 4), b(2, 6) and (2 +`sqrt(3)` ,5) are the vertices of an equilateral triangle
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Solution
The given points are A(2, 4), b(2, 6) and (2 +`sqrt(3)` ,5) Now
`AB =sqrt(((2-2)^2 +(4-6)^2 )) = sqrt((0)^2 +(-2)^2)`
`= sqrt((0+4) =2`
`BC = sqrt((2-2- sqrt(3))^2 + (6-5)^2 ) = sqrt((- sqrt(3))^2 +(1)^2)`
`= sqrt(3+1) = 2`
`AC = sqrt((2-2-sqrt(3))^2 + (4-5)^2 ) = sqrt((- sqrt(3))^2 +(-1)^2)`
`= sqrt(3+1) =2`
Hence, the points A(2, 4), b(2, 6) and (2 +`sqrt(3)` ,5) are the vertices of an equilateral triangle
Concept: Area of a Triangle
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