Question

Show that the points (a, a), (–a, –a) and (– √3 a, √3 a) are the vertices of an equilateral triangle. Also find its area.

Answer 1

Let A (a, a), B(–a, –a) and C(– √3 a, √3 a) be the given points. Then, we have

`AB=\sqrt{(-a-a)^{2}+(-a-a)^{2})=sqrt(4a^2+4a^2) = 2sqrt2a`

`BC=sqrt((-sqrt3a+a)^2+(sqrt3a+a)^2)`

`=>BC=sqrt(a^2(1-sqrt3)^2+a^2(sqrt3+1)^2`

`\Rightarrow BC=a\sqrt{1+3-2\sqrt{3}+1+3+2\sqrt{3}}=a\sqrt{8}=2\sqrt{2}a and,\text{ }AC=\sqrt{(-\sqrt{3}a-a^2)+(\sqrt{3}a-a)}^{2}`

`=>AC=sqrt(a^2(sqrt3+1)+a^2(sqrt3-1)^2)`

`\Rightarrow AC=\sqrt{3+1+2\sqrt{3}+3+1-2\sqrt{3}}=a\sqrt{8}=2\sqrt{2}a`

Clearly, we have

AB = BC = AC

Hence, the triangle ABC formed by the given points is an equilateral triangle.
Now,

`Area of ∆ABC = \frac { \sqrt { 3 } }{ 4 } (side)^2`

`⇒ Area of ∆ABC = \frac { \sqrt { 3 } }{ 4 } × AB^2`

`⇒ Area of ∆ABC = \frac { \sqrt { 3 } }{ 4 } × (2√2 a)^2 sq. units = 2√3 a^2 sq. units`