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# Show that the Points A(1, 2), B(1, 6), C ( 1 + 2 √ 3 , 4 ) Are Vertices of an Equilateral Triangle. - Geometry

Sum

Show that the points A(1, 2), B(1, 6), $C\left( 1 + 2\sqrt{3}, 4 \right)$ are vertices of an equilateral triangle.

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#### Solution

The given points are A(1, 2), B(1, 6), $C\left( 1 + 2\sqrt{3}, 4 \right)$.

$AB = \sqrt{\left( 1 - 1 \right)^2 + \left( 2 - 6 \right)^2} = \sqrt{0 + \left( - 4 \right)^2}$

$= \sqrt{16}$

$= 4$

$BC = \sqrt{\left( 1 - 1 - 2\sqrt{3} \right)^2 + \left( 6 - 4 \right)^2}$

$= \sqrt{12 + 4} = \sqrt{16}$

$= 4$

$AC = \sqrt{\left( 1 - 1 - 2\sqrt{3} \right)^2 + \left( 2 - 4 \right)^2}$

$= \sqrt{12 + 4}$

$= \sqrt{16}$

$= 4$

AB = BC = AC = 4
Since, all the sides of the triangle formed by the points A, B and C are equal so, the tiangle formed is equilateral triangle.

Concept: Standard Forms of Equation of a Line
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#### APPEARS IN

Balbharati Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board
Chapter 5 Co-ordinate Geometry
Practice Set 5.1 | Q 8 | Page 108
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