Question

Prove that the point A(2,4) B(2,6) C (2+,√3,5) are the vertices of an equilateral triangle

Answer 1

this is the answer of your question

Answer 2

$\bigtriangleup ABC$ an equilateral triangle where all angle are equal to $60$ degree.

Step-by-step explanation:

Given;

$A(2,4)$ , $B(2,6)$ and $C(2+\sqrt{3},5)$

∴Side $AB=\sqrt{(y_{2}- y_{1} )^{2}+ (x_{2}- x_{1} )^{2}}$

⇒$AB=\sqrt{(6-4)^{2}+(2-2)^{2} }$

⇒$AB=\sqrt{2^{2} }$

⇒$AB=2$

And, side $BC=\sqrt{(y_{3}- y_{2} )^{2}+ (x_{3}- x_{2} )^{2}}$

⇒$BC=\sqrt{(5- 6 )^{2}+ (2+\sqrt{3} - 2)^{2}}$

⇒$BC=\sqrt{(- 1 )^{2}+ (\sqrt{3} )^{2}}$

⇒$BC=\sqrt{1+3}$

⇒$BC=2$

Also, side $AC=\sqrt{(y_{3}- y_{1} )^{2}+ (x_{3}- x_{1} )^{2}}$

⇒$AC=\sqrt{(5-4)^{2}+ (2+\sqrt{3}-2 )^{2}}$

⇒$AC=\sqrt{(1)^{2}+ (\sqrt{3})^{2}}$

⇒$AC=\sqrt{1+3}$

⇒$AC=2$

∴$AB=BC=AC$

So, If we make $\bigtriangleup ABC$ by point $A$,$B$ and $C$.

Then;

$\bigtriangleup ABC$ an equilateral triangle where all angle are equal to $60$ degree.