Question

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

Answer 1

Answer: yes it is an isosceles triangle

Step-by-step explanation:

Let the vertices be A(5,-2); B(6,4) and C(7,-2)

For the triangle to be an isosceles triangle, 2 of the sides should be equal to each other in magnitude

This implies that either AB = BC (or) BC = AC (or) AB = AC

We first need to find the length of the sides

Formula for finding the length of the side is

$\sqrt{ (x2 - x1)^{2} + (y2 - y1)^{2} }$

Where (x1, y1) and (x2, y2) are the points between which the distance is to be found

Substituting values of the vertices and solving, we get

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

=> AB = $\sqrt{ 1^{2} +6^{2} } = \sqrt{1+36} = \sqrt{37}$

=> AB = $\sqrt{37}$

Similarly while solving by substituting, we get

BC = $\sqrt{37}$

And

AC = $\sqrt{2^{2}} = 2$

Therefore, AB = BC

Thus ABC is an isosceles triangle with AB = BC

Answer 2

Answer:

Put:     A = ( 5, -2 )        B = ( 6, 4 )        C = ( 7, -2 )

Method 1   ---  Calculating lengths

AB = √( (6 - 5)² + (4 - -2)² ) = √( 1² + 6² ) = √37

BC = √( (6 - 7)² + (4 - -2)² ) = √( 1² + 6² ) = √37

As AB = BC, the triangle ABC is isosceles