Question

Show that the points $(a, a), (- a, - a)$ and $(- \sqrt{3} a, \sqrt{3} a)$ are the vertices of an equilateral triangle.

Accepted Answer

Given points are $A (a, a), B (- a, - a) and C (- \sqrt{3} a, \sqrt{3} a)$

Distance formula between the points $A (x_{1}, y_{1}), B (x_{2}, y_{2})$ is $= \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

By applying the distance formula we get,

$A B = \sqrt{{(a - (- a))}^{2} + {(a - (- a))}^{2}}$

$= \sqrt{{(a + a)}^{2} + {(a + a)}^{2}}$

$= \sqrt{{(2 a)}^{2} + {(2 a)}^{2}} = \sqrt{4 a^{2} + 4 a^{2}}$

$= \sqrt{8 a^{2}} = 2 \sqrt{2} a$

$B C = \sqrt{{(- a - (- \sqrt{3} a))}^{2} + (- a - \sqrt{3} a)^{2}}$

$= \sqrt{{(- a + \sqrt{3} a)}^{2} + (a + \sqrt{3} a)^{2}}$

$= \sqrt{{a^{2} (\sqrt{3} - 1)}^{2} + a^{2} (\sqrt{3} + 1)^{2}}$

$= \sqrt{a^{2} ({(\sqrt{3})}^{2} + 1^{2} - 2 \sqrt{3} + {(\sqrt{3})}^{2} + 1^{2} + 2 \sqrt{3})}$

$= \sqrt{a^{2} (3 + 1 - 2 \sqrt{3} + 3 + 1 + 2 \sqrt{3})}$

$= \sqrt{8 a^{2}} = 2 \sqrt{2} a$

$A C = \sqrt{{(a - (- \sqrt{3} a))}^{2} + (a - \sqrt{3} a)^{2}}$

$= \sqrt{{(a + \sqrt{3} a)}^{2} + (a - \sqrt{3} a)^{2}}$

$= \sqrt{{a^{2} (\sqrt{3} + 1)}^{2} + a^{2} (1 - \sqrt{3})^{2}}$

$= \sqrt{8 a^{2}} = 2 \sqrt{2} a$

$A B = B C = A C$

Length of the all sides are equal in an equilateral triangle

Hence, given points are the vertices of an equilateral triangle.

M L Aggarwal Solutions for Chapter: Coordinate Geometry, Exercise 2: Execise 6.2

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Coordinate Geometry, Exercise 2: Execise 6.2

Questions from M L Aggarwal Solutions for Chapter: Coordinate Geometry, Exercise 2: Execise 6.2 with Hints & Solutions

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