Question

Show that the points A( -1, 2) , B ( 3, -1) and C (2 , 6 ) are the vertices of an isosceles right angled triangle​

Answer 1

Given : A ( -1, 2) , B( 3, -1) , C(2, 6)

To prove ,

If the given points are vertices of an isosceles right angled triangle .

Solution ,

If , two sides of a triangle are equal , it is said to be an isosceles triangle .

Using distance formula , we can calculate the distance between any 2 points in the plane.

Hence,

AB = $\sqrt{(3 - (-1))^{2} + (-1 -2)^{2} }$

     = $\sqrt{(16) + (9)}$

     = 5 units

BC = $\sqrt{(2 -3)^{2} + (6 - (-1)^{2} ) }$

     = $\sqrt{(1) + (49)}$

     = 7.1 units (approx.)

AC = $\sqrt{(2- (-1))^{2} + (6 -2)^{2} }$

     = $\sqrt{(9) + (16)}$

     = 5 units

Therefore , AB = AC ≠ BC

Also using Pythagoras Theorem

        $AB^{2} + AC^{2} = BC^{2}$

         $(5)^{2} + (5)^{2} = (7.1)^{2}$

Final answer :

Hence , ABC is an isosceles right angled triangle , right angled at A.

Answer 2

Answer is above in the image☝️