Question

show that the point (a,a),(-a, -a) and -3a,3a)are the vertics of an equilateral triangle
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Answer 1

Correction in the question: Show that the points (a, a),(-a, -a) and (-√3a, √3a) are the vertices of an equilateral triangle .

We've been given three coordinates, and we're asked to show if these three points when joined form an Equilateral triangle.

For three points to form an equilateral triangle, all three sides need to be equal. Therefore, let's try to find the distances of all the three lines and compare them.

Let us name (a, a), (-a, -a) and (-√3a, √3a) as A, B and C respectively.

We'll find the distances of AB, BC and CD using the distance formula.

$\boxed{\sf \Longrightarrow Distance \ Formula = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \ }$

AB's distance:

x₁ = a

x₂ = -a

y₁ = a

y₂ = -a

$\Longrightarrow \sf AB = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\ \\ \\\sf \Longrightarrow AB = \sqrt{(a - (-a))^2 + (a - (-a))^2}\\ \\ \\\sf \Longrightarrow AB = \sqrt{(a + a)^2 + (a + a)^2}\\ \\ \\\sf \Longrightarrow AB = \sqrt{(2a)^2 + (2a)^2}\\ \\ \\\sf \Longrightarrow AB = \sqrt{4a^2 + 4a^2}\\ \\ \\\sf \Longrightarrow AB = \sqrt{8a^2} \ sq.units$

BC's distance:

x₁ = -a

x₂ = -√3a

y₁ = -a

y₂ = √3a

$\Longrightarrow \sf BC = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\ \\ \\\sf \Longrightarrow BC = \sqrt{(-a - (-\sqrt{3}a))^2 + (-a - \sqrt{3}a)^2}\\ \\ \\\sf \Longrightarrow BC = \sqrt{(-a + \sqrt{3}a)^2 + (-a - \sqrt{3}a)^2}\\ \\ \\\sf \Longrightarrow BC = \sqrt{(-a)^2 + (\sqrt{3}a)^2 + 2(-a)(\sqrt{3}a) + (-a)^2 + (\sqrt{3}a)^2 - 2(-a)(\sqrt{3}a)}\\ \\ \\\sf \Longrightarrow BC = \sqrt{a^2 + 3a^2 - 2\sqrt{3}a^2 + a^2 + 3a^2 + 2\sqrt{3}a^2}\\ \\ \\\sf \Longrightarrow BC = \sqrt{8a^2} \ sq.units.$

Ca's distance:

x₁ = -√3a

x₂ = a

y₁ = √3a

y₂ = a

$\Longrightarrow \sf CA = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\ \\ \\\sf \Longrightarrow CA = \sqrt{(-\sqrt{3}a - a)^2 + (\sqrt{3}a - a)^2}\\ \\ \\\sf \Longrightarrow CA = \sqrt{(-\sqrt{3}a)^2 + (a)^2 - 2(-\sqrt{3}a)(a) + (\sqrt{3}a)^2 + (a)^2 - 2(\sqrt{3}a)(a)}\\ \\ \\\sf \Longrightarrow CA = \sqrt{3a^2 + a^2 + 2\sqrt{3}a^2 + 3a^2 + a^2 - 2\sqrt{3}a^2}\\ \\ \\\sf \Longrightarrow CA = \sqrt{8a^2} \ sq.units$

On observing AB, BC, and CA we can say that all three sides are equal to √8a².

∴ ABC forms an equilateral triangle.