Question

EXAMPLE 10 Show that the points (a, a),(-a, - a) and (-3a, 3a) are the vertices of an equilateral triangle. Also find its area.​

Answer 1

Answer:

The triangle is not equilateral.

Step-by-step explanation:

Distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Distance from (a, a) to (-a, -a):

$d_1 = \sqrt{(-a - a)^2 + (-a - a)^2}$

$d_1 = \sqrt{(-2a)^2 + (-2a)^2}$

$d_1 = \sqrt{8a^2}$

Distance from (-a, -a) to (-3a, 3a):

$d_2 = \sqrt{(-3a - (-a))^2 + (3a - (-a))^2}$

$d_2 = \sqrt{(-2a)^2 + (4a)^2}$

$d_2 = \sqrt{20a^2}$

Distance from (-3a, 3a) to (a, a):

$d_3 = \sqrt{(-3a - a)^2 + (3a - a)^2}$

$d_3 = \sqrt{(-4a)^2 + (2a)^2}$

$d_3 = \sqrt{20a^2}$

The three sides do not have the same length, so the triangle is not equilateral.