Question

show that the points form an equaliteral traingle A(a,0)B(-a,o)C(0,a√3)
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Answer 1

Answer:

use the distance formula

=√(x2-x1)^2 + (y2-y1)^2

find distance of AB, BC, AC and prove them equal to each other

Answer 2

GIVEN:

• A = (a , 0), B = (- a , 0), C = (0 , a√3)

TO PROVE:

• The given points form an equaliteral traingle.

PROOF:

• In an equaliteral traingle, all sides are of equal length.

• So we want to prove that AB = BC = CA.

$\boxed{ \bold{\large{ \gray {Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}}}}$

Distance from A to B:

A = (a , 0), B = (- a , 0)

$\sf x_1 = a, \ x_2 = -a$

$\sf y_1 = 0, \ y_2 = 0$

$\sf AB= \sqrt{(-a -a)^2 + (0-0)^2}$

$\sf AB= \sqrt{(-2a)^2 + (0)^2}$

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

$\sf AB= 2a \ units$

Distance from B to C:

B = (- a , 0), C = (0 , a√3)

$\sf x_1 = - a, \ x_2 =0$

$\sf y_1 = 0, \ y_2 = a \sqrt{3}$

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

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

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

$\sf BC= \sqrt{4a^2}$

$\sf BC= 2a \ units$

Distance from C to A:

C = (0 , a√3), A = (a , 0)

$\sf x_1 = 0, \ x_2 =a$

$\sf y_1 = a \sqrt{3}, \ y_2 = 0$

$\sf CA= \sqrt{(a - 0 )^2 + (0 - a \sqrt{3} {)}^{2} }$

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

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

$\sf CA= \sqrt{4a^2 }$

$\sf CA=2a \ units$

AB = BC = CA = 2a units.

The given points form an equilateral triangle.

HENCE PROVED.