Show that the points (2,0), (-2,0) and (0,2) are the vertices of
a triangle. Also state with reason the type of the triangle .

Explanation

Let A(2,0), B(-2,0) and C(0,2)
By distance formula
`AB = sqrt((x_(2) - x_(1))^(2) + (Y_(2) - y_(2))^(2))`
` = sqrt((-2-2)^(2) + (0- 0)^(2))`
` = sqrt((-4)^(2) + (0)^(2))`
` sqrt(16)`
` = 4 `
Similary , ` BC = 2sqrt(2)" and " AC = 2sqrt(2)`
For any triangle
Sum of two sides will always be greter then the third side
` therefore 2sqrt(2) + 2sqrt(2) gt 4 `
i.e. ` 4sqrt(2) gt 4 ` and also ` 4 + 2 sqrt(2) gt 2sqrt(2)`
` therefore A ,B C ` are the vertices of a triangle Now,
` BC^(2) + AC^(2) = (2sqrt(2)^(2)+ 2sqrt(2))^(2)`
` therefore BC^(2) + AC^(2) = 16` ...(1)
`AB^(2) = 4^(2)`
` therefore AB^(2) = 16` ...(2)
` therefore BC^(2) + AC^(2) = AB^(2)" "` [from (1) and (2)]
` therefore ` by converse, of Pythagoras theorem
` Delta ABC ` is a right angled triangle and also
BC = AC
` therefore Delta ABC ` is an isosceles right angled triangle.