Question

Show that the points A(1,2) , B (1,6) , C (1+2root 3,4) are vertices of equilateral triangle.
______

⚡ No irrelevant answers​

Answer 1

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Answer 2

$\large\underline{\sf{Solution-}}$

Given coordinates of triangle ABC are

$\rm :\longmapsto\:Coordinates \: of \: A = (1,2)$

$\rm :\longmapsto\:Coordinates \: of \: B = (1,6)$

$\rm :\longmapsto\:Coordinates \: of \: C = (1 + 2 \sqrt{3} ,4)$

In order to show that points A, B and C are the vertices of equilateral triangle, we have to show that 3 sides are equal.

That means, AB = BC = CA

Where,

AB means distance between A and B coordinate.

BC means distance between B and C coordinate.

CA means distance between A and C coordinate.

Now, we know that,

Distance Formula :-

Let us consider a line segment joining the points A (x₁ , y₁ ) and B (x₂ , y₂), then distance between A and B is given by

$\boxed{ \sf{ AB = \: \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} }}}$

So,

We have

$\rm :\longmapsto\:Coordinates \: of \: A = (1,2)$

$\rm :\longmapsto\:Coordinates \: of \: B = (1,6)$

Thus,

$\rm :\longmapsto\:AB = \sqrt{ {(1 - 1)}^{2} + {(6 - 2)}^{2} }$

$\rm :\longmapsto\:AB = \sqrt{ {(0)}^{2} + {(4)}^{2} }$

$\bf\implies \:AB = 4 \: units$

Now, we have

$\rm :\longmapsto\:Coordinates \: of \: B = (1,6)$

$\rm :\longmapsto\:Coordinates \: of \: C = (1 + 2 \sqrt{3} ,4)$

Thus,

$\rm :\longmapsto\:BC = \sqrt{ {(1 + 2 \sqrt{3} - 1)}^{2} + {(6 - 4)}^{2} }$

$\rm :\longmapsto\:BC = \sqrt{ {( 2 \sqrt{3})}^{2} + {(2)}^{2} }$

$\rm :\longmapsto\:BC = \sqrt{12 + 4}$

$\rm :\longmapsto\:BC = \sqrt{16}$

$\bf\implies \:BC = 4 \: units$

Now, we have

$\rm :\longmapsto\:Coordinates \: of \: A = (1,2)$

$\rm :\longmapsto\:Coordinates \: of \: C = (1 + 2 \sqrt{3} ,4)$

Thus,

$\rm :\longmapsto\:AC = \sqrt{ {(1 + 2 \sqrt{3} - 1)}^{2} + {(4 - 2)}^{2} }$

$\rm :\longmapsto\:AC = \sqrt{ {( 2 \sqrt{3})}^{2} + {(2)}^{2} }$

$\rm :\longmapsto\:AC = \sqrt{12 + 4}$

$\rm :\longmapsto\:AC = \sqrt{16}$

$\bf\implies \:AC = 4 \: units$

Thus, from above we concluded that AB = BC = AC

So, triangle ABC is equilateral.

Hence, points form the vertices of equilateral triangle.

Additional Information :-

Section Formula :-

Let us consider a line segment joining the points A (x₁ , y₁ ) and B (x₂ , y₂) and Let C (x, y) be any point on AB which divides AB internally in the ratio m : n, then coordinates of C is

$\boxed{ \sf{ \:\bf \:( x, y) = \bigg(\dfrac{mx_2 + nx_1}{m + n} , \dfrac{my_2 + ny_1}{m + n} \bigg)}}$

Midpoint Formula :

Let us consider a line segment joining the points A (x₁ , y₁ ) and B (x₂ , y₂) and Let C (x, y) be mid - point of AB, then coordinates of C is

$\boxed{ \sf{ \:\bf \:( x, y) = \bigg(\dfrac{x_2 + x_1}{2} , \: \dfrac{y_2 + y_1}{2} \bigg)}}$