Question

Show that ( 2, -2 ) , ( -2,1 ) and ( 5,2) are the vertices of a right angled triangle ?

( Non-sense answers will be deleted on the spot ! )

( Solution should make sense ! )

Answer 1

Let the given points be A(2,-2), B(-2,1) and C(5,2) are the vertices of a triangle.

Let us calculate the length of each side.

$AB = \sqrt{2 - (-2)^2 + (-2 - 1)^2}$

              = $\sqrt{(4)^2 + (-3)^2}$

              = $\sqrt{16 + 9}$

              = $\sqrt{25}$

              = 5.


$BC = \sqrt{(-2 - 5)^2 + (1 - 2)^2}$

             $= \sqrt{(-7)^2 + (-1)^2}$

            $\sqrt{49 + 1}$

           $\sqrt{50}$

            $5\sqrt{2}$



$CA = \sqrt{(5 - 2)^2 + (2 - (-2))^2}$

              $\sqrt{(3)^2 + (4)^2}$

              $\sqrt{16 + 9}$

             $\sqrt{25}$

             5.



Then BC is the longest side.

BC^2 = AC^2 + AB^2

$( 5\sqrt{2})^2 = 5^2 + 5^2$

50 = 25 + 25.

50 = 50.



Therefore ABC is a right-angled triangle.


Hope this helps!  ----------------- Good Luck.

Answer 2

Hi

Here is your answer,

AB = √(x₂ - x₁)² + (y₂ - y₁)²

       = √(-2-2)² + (1-(-2)²

      = √(4)² + (1 + 3)²

      = √ 16 + 9  = √25 = 5

BC = √(x₂ - x₁)² + (y₂ - y₁)²

      = √( 5 -(-2)² + (2 -1)²

      = √( 5 + 2)² + (1)²

      = √(7)² + (1)²

      = √ 49 + 1 

      = √50

      = 5√2

CA = √(x₂ - x₁)² + (y₂ - y₁)²

      = √( 2 - 5)² + ( -2-2)²

      = √(-3)² + (4)²

      = √ 9 + 16

      = √25

      = 5                                                                { ∴ BC IS THE LONGEST SIDE }

BC² = AC² + AB²

(5√2)² = (5)² + (5)²

50 = 25 + 25

50 = 50


∴ ABC is right angled traingle

Hope it helps you !