Question

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

Answer 1

Let they be

• A(a,a)
• B(-a,-a)
• C(-√3a,√3a)

Find distances

`\\ \rm\Rrightarrow AB=√((a+a)^2+(a+a)^2)=√(4a^2+4a^2)=√(8a^2)=2√(2)a`

`\\ \rm\Rrightarrow BC=√((-√3a+a)^2+(√3a+a)^2)`

• (a+b)^2+(a-b)^2=2a²+2b²

`\\ \rm\Rrightarrow BC=\sqrt{2(√(3)a)^2+2a^2}=√(6a^2+2a^2)=√(8a^2)=2√(2)a`

`\\ \rm\Rrightarrow AC=\sqrt{(a+√(3)a)^2+(a-√(3)a)^2}=2√(2)a`

As

• AB=AC=BC

The traingle is equilateral

Area

`\\ \rm\Rrightarrow (√(3))/(4)a^2`

`\\ \rm\Rrightarrow (√(3))/(4)(2√(2)a)^2`

`\\ \rm\Rrightarrow (√(3))/(4)(8a^2)`

`\\ \rm\Rrightarrow 2√(3)a^2`

Answer 2

`{\large{\textsf{\textbf{\underline{\underline{Solution \: :}}}}}}`

Let, the given points be

`\bullet \: \tt A (a, a)`

`\bullet \: \tt B (-a, -a)`

`\bullet \: \tt C ( - √(3)a , √(3)a)`

Now,

Finding the distances of AB, BC and CD using distance formula.

`\star \: \bold{\underline \red{\sf{ Finding \: AB }}}`

Using,

`\tt Distance \: Formula = \sqrt{{(x{\small_(2)} -x{\small_(1)} )}^(2) +{(y{\small_(2)} -y{\small_(1)} )}^(2)}`

where

`\sf x{\small_(2)} = - a`

`\sf x{\small_(1)} = a`

`\sf y{\small_(2)} = - a`

`\sf y{\small_(1)} = a`

Putting the values,

`\longrightarrow \sf AB = \sqrt{( { - a - a)}^(2) + {( - a - a)}^(2) }`

`\longrightarrow \sf AB = \sqrt{( { - 2 a)}^(2) + {( -2 a)}^(2) }`

`\longrightarrow \sf AB = \sqrt{4{ a}^(2) + 4{ a}^(2) }`

`\longrightarrow \sf AB = \sqrt{8{ a}^(2) }`

`\longrightarrow \sf AB = \sqrt{2 * 2 * 2 \: { a}^(2) }`

`\longrightarrow \sf AB = \red{2 √( 2 )a}`

`\star \: \bold{\underline \green{\sf{ Finding \: BC }}}`

Using,

`\tt Distance \: Formula = \sqrt{{(x{\small_(2)} -x{\small_(1)} )}^(2) +{(y{\small_(2)} -y{\small_(1)} )}^(2)}`

where

`\sf x{\small_(2)} = - √(3)a`

`\sf x{\small_(1)} = - a`

`\sf y{\small_(2)} = √(3)a`

`\sf y{\small_(1)} = - a`

Putting the values,

`\longrightarrow \sf BC = \sqrt{ \bigg[ { - √(3)a - (- a) \bigg]}^(2) + {\bigg[ √(3)a - (- a)\bigg]}^(2) }`

`\longrightarrow \sf BC = \sqrt{ { ( - √(3)a + a) }^(2) + {( √(3) a + a)}^(2) }`

Using (a + b)² = a² + 2(a)(b) + b².

`\longrightarrow \sf BC = \sqrt{ { (- √(3)a) }^(2) + 2( - √(3) a)(a) + {(a)}^(2) + {( √(3) a + a)}^(2) }`

`\longrightarrow \sf BC = \sqrt{ 3 {a}^(2) - 2 √(3) {a}^(2) + {a}^(2) + {( √(3) a + a)}^(2) }`

Again, using (a + b)² = a² + 2(a)(b) + b².

`\longrightarrow \sf BC = \sqrt{ 3 {a}^(2) - 2 √(3) {a}^(2) + {a}^(2) + { ( √(3)a) }^(2) + 2( √(3) a)(a) + {(a)}^(2) }`

`\longrightarrow \sf BC = \sqrt{ 3 {a}^(2) \:\cancel{ - 2 √(3) {a}^(2)} + {a}^(2) + 3 {a}^(2) \:\cancel{ + 2 √(3) {a}^(2)} + {a}^(2) }`

`\longrightarrow \sf BC = \sqrt{ 3 {a}^(2) + {a}^(2) + 3 {a}^(2) + {a}^(2) }`

`\longrightarrow \sf BC = \sqrt{ 8 {a}^(2) }`

`\longrightarrow \sf BC = \sqrt{2 * 2 * 2 \: { a}^(2) }`

`\longrightarrow \sf BC = \green{2 √( 2 )a}`

`\star \: \bold{\underline \orange{\sf{ Finding \: AC }}}`

Using,

`\tt Distance \: Formula = \sqrt{{(x{\small_(2)} -x{\small_(1)} )}^(2) +{(y{\small_(2)} -y{\small_(1)} )}^(2)}`

where

`\sf x{\small_(2)} = - √(3)a`

`\sf x{\small_(1)} = a`

`\sf y{\small_(2)} = √(3)a`

`\sf y{\small_(1)} = a`

Putting the values,

`\longrightarrow \sf AC = \sqrt{ { ( - √(3)a - a)}^(2) + {(√(3)a - a)}^(2) }`

Using (a - b)² = a² - 2(a)(b) + b².

`\longrightarrow \sf AC = \sqrt{ {( - √(3) a)}^(2) - 2( - √(3) a)(a) + {(a)}^(2) + {(√(3)a - a)}^(2) }`

`\longrightarrow \sf AC = \sqrt{ 3 {a}^(2) + 2 √(3) {a}^(2) + {a}^(2) + {(√(3)a - a)}^(2) }`

Again, using (a - b)² = a² - 2(a)(b) + b².

`\longrightarrow \sf AC = \sqrt{ 3 {a}^(2) + 2 √(3) {a}^(2) + {a}^(2) + {( √(3) a)}^(2) - 2( √(3)a)(a) + {(a)}^(2) }`

`\longrightarrow \sf AC = \sqrt{ 3 {a}^(2) \: \cancel{ + 2 √(3) {a}^(2) } + {a}^(2) + 3 {a}^(2) \: \cancel{ - 2 √(3) {a}^(2) } + {a}^(2) }`

`\longrightarrow \sf AC = \sqrt{ 3 {a}^(2) + {a}^(2) + 3 {a}^(2) + {a}^(2) }`

`\longrightarrow \sf AC = \sqrt{ 8 {a}^(2) }`

`\longrightarrow \sf AC = \sqrt{2 * 2 * 2 \: { a}^(2) }`

`\longrightarrow \sf AC = \sqrt{2 * 2 * 2 \: { a}^(2) }`

`\longrightarrow \sf AC = \orange{{2√(2 )}a}`

Since,

AB = BC = CA =
`\sf {{2√(2 )}a}`

Therefore, the
`\triangle` ABC formed by the given points is an equilateral triangle.

For,

Finding the area of
`\triangle` ABC

Using,

`\longrightarrow \: \tt Area_((equilateral \: \triangle)) = ( √(3) )/(4) \: {(side)}^(2)`

Putting,

• side =
`\sf {{2√(2 )}a}`

`\longrightarrow \: \tt Area_((equilateral \: \triangle)) = ( √(3) )/(4) \: {(2 √(2)a )}^(2)`

`\longrightarrow \: \tt Area_((equilateral \: \triangle)) = ( √(3) )/(4 ) * 2 * 2 * 2 * {a}^(2)`

`\longrightarrow \: \tt Area_((equilateral \: \triangle)) = \frac{ √(3) }{ \cancel{4}} * \cancel{8 }{a}^(2)`

`\longrightarrow \: \tt Area_((equilateral \: \triangle)) = \purple{2 √(3) {a}^(2) \: sq. \: units}`

`\therefore \sf Area = 2 √(3) {a}^(2) \: sq. \: units`

`{\underline{\rule{310pt}{2pt}}}`