Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Let the points (5, - 2), (6, 4), and (7, - 2) represent the vertices A, B, and C of the given triangle

We know that the distance between the two points is given by the Distance Formula,

Distance Formula =  √[(x₂ ^_-  x₁)² + (y₂ - y₁)²]

To find AB, that is distance between points A (5, - 2) and B (6, 4), let x₁ = 5, y₁ = -2, x₂ = 6, y₂ = 4

AB = √[( 6 - 5 )² + (4 - (-2))²]

= √[(1)² + (6)²]

= √1 + 36

= √37

To find BC, distance between Points B (6, 4) and C (7, - 2), let x₁ = 6, y₁ = 4, x₂ = 7, y₂ = - 2

BC = √ [( 7 - 6 )² + (-2 - 4)²]

= √[(1)² + (- 6)²]

= √1 + 36

= √37

To find AC, that is distance between Points A (5, - 2) and C (7, - 2), let x₁ = 5, y₁ = - 2, x₂ = 7, y₂ = - 2

AC = √ [( 7 - 5 )² + (-2 - (- 2))²]

= √[(2)² + (0)²]

= 2

From the above values of AB, BC and AC we can conclude that AB = BC. As the two sides are equal in length, therefore, ABC is an isosceles triangle.