Question

show that the points (-1,-1),(1,1)&(-√3,√3)are the vertices of an equilateral triangle

Answer 1

Answer:

Step-by-step explanation:

Let the points be A(-1,-1),B(1,1)&C(-√3,√3)

$AB=\sqrt{ {({x_1}-{x_2})}^2+{({y_1}-{y_2})}^2}}\\=\sqrt{ {(-1-1)}^2+{(-1-1)}^2}}\\=\sqrt{4+4}\\=\sqrt{8}=2\sqrt{2}$

$BC=\sqrt{ {({x_1}-{x_2})}^2+{({y_1}-{y_2})}^2}}\\=\sqrt{ {(1+\sqrt{3})}^2+{(1-\sqrt{3})}^2}}\\=\sqrt{1+3+2\sqrt{3}+1+3-2\sqrt{3}}\\=\sqrt{8}=2\sqrt{2}$

$AC=\sqrt{ {({x_1}-{x_2})}^2+{({y_1}-{y_2})}^2}}\\=\sqrt{ {(-1+\sqrt{3})}^2+{(-1-\sqrt{3})}^2}}\\=\sqrt{1+3-2\sqrt{3}+1+3+2\sqrt{3}}\\=\sqrt{8}=2\sqrt{2}$

This implies

AB=BC=AC

Hence the given points form an equilateral triangle.

Answer 2

we know, any triangle is said to be equilateral triangle when all sides of triangle be equal.

so, use distance formula and find length of all sides of given triangle.
if (x1, y1) and (x2, y2) two points are given then, distance between them = √{(x1 - x2)² +(y1 - y2)²}

Let A = (-1, -1) , B = (1, 1) and C = (-√3, √3)

length of side AB = √{(-1 - 1)² + (-1 - 1)²}
= √{(-2)² + (-2)²}
= √{4 + 4} = √8
=2√2

length of side BC = √{(-1 + √3)² + (-1 - √3)²}
= √{√3² + 1² - 2√3 + √3² + 1² + 2√3}
= √{3 + 1 + 3 + 1}
= √{8}
= 2√2

length of side CA = √{(-√3 + 1)² + (√3 + 1)²}
= √{√3² + 1² - 2√3 + √3² + 1² + 2√3}
= √{3 + 1 + 3 + 1 }
= √8
= 2√2

here we see that length of side AB = length of side BC = length of side CA

so, ABC is an equilateral triangle.