Prove that the triangle with vertices with (3,5), (-2,6), and (1,3) is a right triangle.

If the sum of square of distance between two vertices is equal to the square of distance between third vertices, then the triangle is a right angled triangle.

By above definition, we get

[tex]BC^{2} + CA^{2} = AB^{2}[/tex] ----- eqn 1

Where AB is the distance between vertices A and B

BC is the distance between vertices B and C

CA is the distance between vertices C and D

Distance between any two vertices of a triangle is given as

[tex]Distance = \sqrt{\left(x_{2} -x_{1}\right )^{2} + \left(y_{2} -y_{1}\right)^{2} }[/tex] ------- eqn 2

Step 1:

Let us find the distance between the vertices A(3,5) and B(-2,6)

By using equation 2, we get

[tex]x_{1} = 3, x_{2} = -2, y_{1} = 5 \text { and } y_{2} = 6[/tex]

Distance between vertices A and B =[tex]\sqrt{(-2-3)^{2}+(6-5)^{2}}[/tex]

= [tex]\sqrt{(-5)^{2} + (1)^{2}}[/tex]

= [tex]\sqrt{25+1}[/tex]

= [tex]\sqrt{26} units[/tex]

Step 2:

Let us find the distance between the vertices B(-2,6) and C(1,3)

By using equation 2, we get

[tex]x_{1} = -2, x_{2} = 1, y_{1} = 6 \text { and } y_{2} = 3[/tex]

Distance between vertices B and C = [tex]\sqrt{(1-(-2))^{2}+(3-6)^{2}}[/tex]

= [tex]\sqrt{(1-(-2))^{2} + (3-6)^{2}}[/tex]

= [tex]\sqrt{3^{2}+9}[/tex]

= [tex]\sqrt{9+9}[/tex]

= [tex]\sqrt{18} \text { units }[/tex]

Step 3:

Let us find the distance between the vertices C(1,3) and D(3,5)

By using equation 2, we get

[tex]x_{1} = 1, x_{2} = 3, y_{1} = 3 \text { and } y_{2} = 5[/tex]

Distance between the vertices C and A

= [tex]\sqrt{(3-1)^{2} + (5-3)^{2}}[/tex]

= [tex]\sqrt{2^{2} + 2^{2}}[/tex]

= [tex]\sqrt{4+4}[/tex]

= [tex]\sqrt{8} units[/tex]

Step 4:

By using equation 1,

[tex]BC^{2} + CA^{2} = AB^{2}[/tex]

[tex](\sqrt{18})^{2} + (\sqrt{8})^{2} = (\sqrt{26})^{2}[/tex]

18 + 8 = 26

26 = 26

Hence the condition is satisfied. So the given triangle with vertices with (3,5), (-2,6), and (1,3) is a right triangle.