Question

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

Maths, class 10.​

Answer 1

Given :

•The points of the given triangle :

•A=(a, a)

•B=(-a, -a)

•C=(-√3a, √3a)

To Find :

We have to show that the triangle with vertices A(a,a) , B(-a, -a) ,C(-√3a, √3a) is an equilateral triangle =?

Solution :

Let A(a,a) ,B(-a, -a) ,C(-√3a, √3a) be the points of triangle.

Now ,the length of AB=

$\\ \implies \sf \: \sqrt{ {(a - ( - a))}^{2} + {(a - ( - a))}^{2} }$

$\implies \sf \sqrt{8 {a}^{2} }$

$\implies \sf \: 2 \sqrt{2} a\: units$

Now, the length of BC=

$\\ \implies \sf \: \sqrt{ {( - a + \sqrt{3a} )}^{2} + {( - a - \sqrt{3a} )}^{2} }$

$\implies \sf \: \sqrt{2( {a}^{2} + {( \sqrt{3}a) }^{2}) }$

$\implies \sf \: 2 \sqrt{2} a \: units$

Since we know that,

$\\ \red \bigstar \boxed{ \bf{ {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )}}$

Again, the length of CA

$\\ \implies \sf \: \sqrt{ {(a + \sqrt{3}a) }^{2} + {(a - \sqrt{3}a) }^{2} }$

$\implies \sf \: \sqrt{2( {a}^{2} + {( \sqrt{3}a) }^{2} ) }$

$\implies \sf \: 2 \sqrt{2} a \: units$

$\red\bigstar\boxed{\sf{AB=BC=CA=2 \sqrt{2} a \: units}}$

Therefore, all the sides of the triangle are equal in length and hence the triangle is an equilateral triangle.