Question

show that points A(a,0) B(-a,0) C(0,a√3) form

equilateral triangle​

Answer 1

Given Question :-

• Show that points A(a,0) B(-a,0) C(0,a√3) form an equilateral triangle.

ANSWER

GIVEN :-

• Points A(a,0) B(-a,0) C(0,a√3)

TO SHOW :-

• Points A(a,0) B(-a,0) C(0,a√3) forms equilateral triangle.

FORMULA USED :-

DISTANCE FORMULA

We know that,

Distance between two points is calculated by using the formula given below,

$\longmapsto \rm \: \rm D = \sqrt{ {(x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2} }$

CALCULATION :-

DISTANCE BETWEEN A and B

Here,

• • x₁ = a

• • x₂ = - a

• • y₁ = 0

• • y₂ = 0

So, distance between these points is,

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

$\longmapsto \rm \: AB = \sqrt{ {(2a)}^{2} }$

$\longmapsto \boxed{ \green{ \bf \: AB \: = \: 2a}}$

DISTANCE BETWEEN B and C

Here,

• • x₁ = - a

• • x₂ = 0

• • y₁ = 0

• • y₂ = a√3

So, distance between these points is,

$\longmapsto \rm \: BC = \sqrt{ {(0 + a)}^{2} + {( \sqrt{3}a - 0) }^{2} }$

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

$\longmapsto \rm \: BC = \sqrt{ {4a}^{2} }$

$\longmapsto \boxed{ \green{ \bf \:BC \: = \: 2a }}$

DISTANCE BETWEEN A and C

Here,

• • x₁ = a

• • x₂ = 0

• • y₁ = 0

• • y₂ = a√3

So, distance between these points is,

$\longmapsto \rm \: AC = \sqrt{ {(0 - a)}^{2} + {( \sqrt{3}a - 0) }^{2} }$

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

$\longmapsto \rm \: AC = \sqrt{4 {a}^{2} }$

$\longmapsto \boxed{ \green{ \bf \: AC \: = \: 2a}}$

$\longmapsto \boxed{ \pink{ \bf \: Hence, \: AB = BC = AC}}$

$\longmapsto \boxed{ \red{ \bf \: Hence, \: \triangle \: ABC \: is \: equilateral.}}$