Question

check whether (5 -2) (6 4) and (7 -2) are the vertices of an isosceles triangle

Answer 1

Answer:

(5 -2) (6 4) and (7 -2) are the vertices of an isosceles triangle

Step-by-step explanation:

Let Say

A ( 5 , -2)

B (6 ,  4)

C ( 7 , -2)

Length of each sides

$AB =\sqrt{(6-5)^2 + (4 -(-2))^2}  = \sqrt{1^2 + 6^2}  = \sqrt{1 + 36}  = \sqrt{37}$

$AC =\sqrt{(7-5)^2 + (-2 -(-2))^2}  = \sqrt{2^2 + 0^2}  = \sqrt{4 + 0}  = \sqrt{4} = 2$

$BC =\sqrt{(7-6)^2 + (-2 -4)^2}  = \sqrt{1^2 + (-6)^2}  = \sqrt{1 + 36}  = \sqrt{37}$

AB = BC = √37

Hence verified that these vertices are of an isosceles triangles

Answer 2

Answer:

Step-by-step explanation:

We can check the same by application of 'distance formula' which is used to find the distance between two points whose coordinates are known.

If A(x₁, y₁) and B (x₂, y₂) represent 2 points in same plane, then by distance formula, the distance 'd' between these points is given by

d = √ (x₂ - x₁)² + (y₂ - y₁)²

Let the given points be written as:

A (5,-2)

B (6, 4)

C (7, -2)

AB = √(6 - 5)² + (4+2)² = √( 1 + 36) =  √37 units

BC = √(7 - 6)² + (-2-4)² = √(1 + 36) = √37 units

CA = √(7-5)² + (-2+2)² = √(4+0) = 2 units

As, AB = BC = √37 units; The given vertices indeed form an isosceles trianlge.