MHB Is this triangle an isosceles triangle?

[mathdad]

Use the distance formula to show that the triangle with the given vertices is an isosceles triangle.

A(0, 2), B(7, 4), C(2, -5)

I must use the distance formula to find AB, BC and AC.
Two sides or lengths must be equal and one side different to be an isosceles triangle.

Correct?

 

[MarkFL]

Yes, the distance formula would be a good way to proceed, but recall, an equilateral triangle is a special case of an isosceles triangle...so, you could have all 3 sides equal in length and still call it an isosceles triangle...much like you can call a square a rectangle that just happens to have all 4 sides being equal in length.

 

[mathdad]

Yes, the distance formula would be a good way to proceed, but recall, an equilateral triangle is a special case of an isosceles triangle...so, you could have all 3 sides equal in length and still call it an isosceles triangle...much like you can call a square a rectangle that just happens to have all 4 sides being equal in length.

I will show my work when time allows.

 

[mathdad]

I will not answer this question using MathMagic Lite.

A(0, 2), B(7, 4), C(2, -5)

AB = sqrt{(7 - 0)^2 + (4 - 2)^2}

AB = sqrt{49 + 4}

AB = sqrt53}

BC = sqrt{(2 - 7)^2 + (-5 - 4)^2}

BC = sqrt{25 + 81}

BC = sqrt{106}

AC = sqrt{(2 - 0)^2 + (-5 - 2)^2}

AC = sqrt{4 + 49}

AC = sqrt{53}

Side AB = side AC.

BC is different than the other two sides of the triangle.

Therefore, triangle ABC is isosceles.