Show that the triangle with vertices A(1,1),B(4,5) and C(0,2) is isosceles?

To show that the triangle with vertices A(1,1), B(4,5), and C(0,2) is isosceles, we need to show that at least two sides of the triangle are equal in length.

First, we calculate the lengths of the three sides of the triangle using the distance formula:

AB = √((4-1)^2 + (5-1)^2) = √(3^2 + 4^2) = √(9 + 16) = √25 = 5

BC = √((0-4)^2 + (2-5)^2) = √((-4)^2 + (-3)^2) = √(16 + 9) = √25 = 5

AC = √((0-1)^2 + (2-1)^2) = √((-1)^2 + 1^2) = √(1 + 1) = √2

Since AB = BC = 5, we can conclude that the triangle with vertices A(1,1), B(4,5), and C(0,2) is isosceles.

Distance of AB =√(4-1)²+(5-1)²
=√(3)²+(4)²
=√9+16
=√25
=5 units

Distance of AC =√(0-1)²+(2-1)²
=√(-1)²+(1)²
=√1+1
=√2
=1.41 units

Distance of BC =√(0-4)²+(2-5)²
=√(-4)²+(-3)²
=√16+9
=√25
=5 units

Therefore AB=BC
Hence the triangle is isosceles