Question

show that (a,a);(-a,a) and (√3a,√3a) are the vertices of equilateral triangle...
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Answer 1

Let A(a, a) , B(-a, a) and C(√3a,√3a)

Here,

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

BC
$= \sqrt{( - a + \sqrt{3a) {}^{2} } + ( - a + \sqrt{3a) {}^{2} } }$
$= a {}^{2} + {3a}^{2} - 2 \sqrt{3 {a}^{2} } + {a}^{2} + 2 \sqrt{3a {}^{2} }$
$= 8 {a}^{2}$
$= 2 \sqrt{2a}$

AC
$= \sqrt{( a + \sqrt{3a) {}^{2} } + ( - a + \sqrt{3a {}^{2} }) }$
$= a {}^{2} + {3a}^{2} + 2 \sqrt{3 {a}^{2} } + {a}^{2} - 2 \sqrt{3a {}^{2} }$
$= 8a {}^{2}$
$= 2 \sqrt{2a}$

Here, AB=BC=AC. Therefore, it is an equilateral triangle.

Answer 2

Show that (a,a);(-a,a) and (√3a,√3a) are the vertices of equilateral triangle.

Question will be,

Show that (a,a);(-a,-a) and (-√3a,√3a) are the vertices of equilateral triangle.

Good question,

Here is your perfect answer!

An equilateral triangle hsv all three sides equal.

AB² = {a-(-a)}² + {a-(-a)}²

= (2a)² + (2a)²

= 4a² + 4a²

= 8a²,

BC² = (-a-(-√3a)² + (-a-√3a)²

= (-a +√3a)² + (-a-√3a)²

= (a² + 3a² - 2√3a² + a² + 3a² + 2√3a²)

= 8a²,

AC² = {a-(-√3a)}² + (a-√3a)²

= (a+√3a)² + (a-√3a)²

= (a² + 3a² + 2√3a² + a² + 3a² - 2√3a²)

= 8a².

Since AB² = BC² = AC²,

AB=BC=AC, proved.