Question

show that A(6,4),B(5,-2) and C(7,-2) are the vertices of an isosceles triangle

Answer 1

Using distance formula

Distance = √(x₂-x₁)² + (y₂-y₁)²


AB² = (6-5)² + (4+2)²
AB² = 37

BC² = (5-7)² + (-2+2)²
BC² = 5

AC² = (6-7) ²+(4+2)²
AC² = 37

Since AB² =AC², so AB =AC
If in a triangle two sides are equal then the triangle is isosceles.

Answer 2

Answer:  The proof is given below.

Step-by-step explanation:  We are given to show that the points  A(6,4),B(5,-2) and C(7,-2) are the vertices of an isosceles triangle.

We know that

a triangle is isosceles if the lengths of any two sides of the triangle are equal.

The lengths of the three sides AB, BC and CA of triangle ABC can be calculated using distance formula as follows :

$AB=\sqrt{(5-6)^2+(-2-4)^2}=\sqrt{1+36}=\sqrt{37}~\textup{units},\\\\\\BC=\sqrt{(7-5)^2+(-2+2)^2}=\sqrt{4+0}=\sqrt{4}=2~\textup{units},\\\\\\CA=\sqrt{(6-7)^2+(4+2)^2}=\sqrt{1+36}=\sqrt{37}~\textup{units}.$

Since the lengths of the sides AB and CA are equal, so the given triangle with vertices A, B and C is an isosceles triangle.

Hence showed.