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

$\\ \rm\Rrightarrow AB=\sqrt{(a+a)^2+(a+a)^2}=\sqrt{4a^2+4a^2}=\sqrt{8a^2}=2\sqrt{2}a$

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

• (a+b)^2+(a-b)^2=2a²+2b²

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

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

As

• AB=AC=BC

The traingle is equilateral

Area

$\\ \rm\Rrightarrow \dfrac{\sqrt{3}}{4}a^2$

$\\ \rm\Rrightarrow \dfrac{\sqrt{3}}{4}(2\sqrt{2}a)^2$

$\\ \rm\Rrightarrow \dfrac{\sqrt{3}}{4}(8a^2)$

$\\ \rm\Rrightarrow 2\sqrt{3}a^2$

Answer 2

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

Let, the given points be

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

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

$\bullet \: \tt C ( - \sqrt{3}a , \sqrt{3}a)$

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,

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

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

$\longrightarrow \sf AB = \sqrt{4{ a}^{2} + 4{ a}^{2} }$

$\longrightarrow \sf AB = \sqrt{8{ a}^{2} }$

$\longrightarrow \sf AB = \sqrt{2 \times 2 \times 2 \: { a}^{2} }$

$\longrightarrow \sf AB = \red{2 \sqrt{ 2 }a}$

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

Using,

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

where

$\sf x{\small_{2}} = - \sqrt{3}a$

$\sf x{\small_{1}} = - a$

$\sf y{\small_{2}} = \sqrt{3}a$

$\sf y{\small_{1}} = - a$

Putting the values,

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

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

Using (a + b)² = a² + 2(a)(b) + b².

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

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

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

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

$\longrightarrow \sf BC = \sqrt{ 3 {a}^{2} \:\cancel{ - 2 \sqrt{3} {a}^{2}} + {a}^{2} + 3 {a}^{2} \:\cancel{ + 2 \sqrt{3} {a}^{2}} + {a}^{2} }$

$\longrightarrow \sf BC = \sqrt{ 3 {a}^{2} + {a}^{2} + 3 {a}^{2} + {a}^{2} }$

$\longrightarrow \sf BC = \sqrt{ 8 {a}^{2} }$

$\longrightarrow \sf BC = \sqrt{2 \times 2 \times 2 \: { a}^{2} }$

$\longrightarrow \sf BC = \green{2 \sqrt{ 2 }a}$

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

Using,

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

where

$\sf x{\small_{2}} = - \sqrt{3}a$

$\sf x{\small_{1}} = a$

$\sf y{\small_{2}} = \sqrt{3}a$

$\sf y{\small_{1}} = a$

Putting the values,

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

Using (a - b)² = a² - 2(a)(b) + b².

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

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

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

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

$\longrightarrow \sf AC = \sqrt{ 3 {a}^{2} \: \cancel{ + 2 \sqrt{3} {a}^{2} } + {a}^{2} + 3 {a}^{2} \: \cancel{ - 2 \sqrt{3} {a}^{2} } + {a}^{2} }$

$\longrightarrow \sf AC = \sqrt{ 3 {a}^{2} + {a}^{2} + 3 {a}^{2} + {a}^{2} }$

$\longrightarrow \sf AC = \sqrt{ 8 {a}^{2} }$

$\longrightarrow \sf AC = \sqrt{2 \times 2 \times 2 \: { a}^{2} }$

$\longrightarrow \sf AC = \sqrt{2 \times 2 \times 2 \: { a}^{2} }$

$\longrightarrow \sf AC = \orange{{2\sqrt{2 }}a}$

Since,

AB = BC = CA = $\sf {{2\sqrt{2 }}a}$

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

For,

Finding the area of $\triangle$ ABC

Using,

$\longrightarrow \: \tt Area_{(equilateral \: \triangle)} = \dfrac{ \sqrt{3} }{4} \: {(side)}^{2}$

Putting,

• side = $\sf {{2\sqrt{2 }}a}$

$\longrightarrow \: \tt Area_{(equilateral \: \triangle)} = \dfrac{ \sqrt{3} }{4} \: {(2 \sqrt{2}a )}^{2}$

$\longrightarrow \: \tt Area_{(equilateral \: \triangle)} = \dfrac{ \sqrt{3} }{4 } \times 2 \times 2 \times 2 \times {a}^{2}$

$\longrightarrow \: \tt Area_{(equilateral \: \triangle)} = \dfrac{ \sqrt{3} }{ \cancel{4}} \times \cancel{8 }{a}^{2}$

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

$\therefore \sf Area = 2 \sqrt{3} {a}^{2} \: sq. \: units$

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