Question

Find the type of triangle formed by point A( a, a ) , B( - a , - a ) and C( - a√3, a√3 ).

Class 10

Coordinate Geometry

Answer 1

We know that Distance between the Points (x₁ , y₁) and (x₂ , y₂) is given by :

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

Let us Find the Distance between Points C and A

here x₁ = $-a\sqrt{3}$ and x₂ = a and y₁ = $a\sqrt{3}$ and y₂ = a

$\implies\sqrt{(-a\sqrt{3} - a)^2\;+(a\sqrt{3} - a)^2$

$\implies\sqrt{(a\sqrt{3} + a)^2\;+(a\sqrt{3} - a)^2$

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

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

Let us Find the Distance between Points C and B

here x₁ = $-a\sqrt{3}$ and x₂ = -a and y₁ = $a\sqrt{3}$ and y₂ = -a

$\implies\sqrt{(-a\sqrt{3} + a)^2\;+(a\sqrt{3} + a)^2$

$\implies\sqrt{(a\sqrt{3} - a)^2\;+(a\sqrt{3} + a)^2$

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

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

Let us find the Distance between A and B

here x₁ = a and x₂ = -a and y₁ = a and y₂ = -a

$\implies\sqrt{(a + a)^2\;+(a + a)^2$

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

As AB = CB = CA, The Given Triangle is an Equilateral Triangle.

Answer 2

Given points are (a,a), B(-a,-a), C(-a√3, a√3).

(i) Calculating AB:

Here, x₁ = a, y₁ = a, x₂ =  -a, y₂ = -a.

⇒ AB = √(x₂ - x₁)^2 + (y₂ - y₁)^2

         = √(-a - a)^2 + (-a - a)^2

         = √(-2a)^2 + (-2a)^2

         = 2√2 a

(ii) Calculating BC:

Here, x₁ = -a, y₁ = -a, x₂ = -a√3, y₂ = a√3.

⇒ BC = √(-a√3 + a)^2 + (a√3 + a)^2

          = √4a^2 - 2√3a^2 + 4a^2 + 2√3a^2

          = √8a^2

          = 2√2 a.

(iii) Calculating AC:

Here, x₁ = a, y₁ = a, x₂ = -a√3, y₂ = a√3

⇒ AC = √(-a√3 - a)^2 + (a√3 - a)^2

          = √4a^2 + 2√3a^2 + 4a^2 - 2√3a^2

          = √8a^2

          = 2√2 a.

Since, all the three sides are equal. Therefore, it forms an equilateral triangle.

Hope it helps!