we are given
three vertices as
A (2,6) B(0,2) C (3,3)
now, we can draw these vertices and find triangle
we can use distance formula
and find AB , BC and AC
[tex] d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} [/tex]
Calculation of AB:
We can use formula
[tex] d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} [/tex]
A (2,6) B(0,2)
[tex] AB=\sqrt{(0-2)^2+(6-2)^2} [/tex]
[tex] AB=\sqrt{20} [/tex]
Calculation of BC:
We can use formula
[tex] d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} [/tex]
B(0,2) C(3,3)
[tex] BC=\sqrt{(3-0)^2+(3-2)^2} [/tex]
[tex] BC=\sqrt{10} [/tex]
Calculation of AC:
We can use formula
[tex] d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} [/tex]
A(2,6) C(3,3)
[tex] AC=\sqrt{(3-2)^2+(3-6)^2} [/tex]
[tex] AC=\sqrt{10} [/tex]
Checking right angled triangle:
We can check it by Pythagoras theorem
[tex] AB^2=BC^2+AC^2 [/tex]
now, we can plug values
[tex] (\sqrt{20})^2=(\sqrt{10})^2+(\sqrt{10})^2 [/tex]
[tex] 20=10+10 [/tex]
[tex] 20=20 [/tex]
we can see that
left side is equal to right side
so, triangle ABC is right angled triangle.........Answer
