Question

The vertices of ABC are A (2,6) B(0,2) C (3,3). What type of triangle is ABC

Answer 1

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Answer 2

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

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Calculation of AB:

We can use formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

A (2,6) B(0,2)

$AB=\sqrt{(0-2)^2+(6-2)^2}$

$AB=\sqrt{20}$

Calculation of BC:

We can use formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

B(0,2) C(3,3)

$BC=\sqrt{(3-0)^2+(3-2)^2}$

$BC=\sqrt{10}$

Calculation of AC:

We can use formula

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

A(2,6) C(3,3)

$AC=\sqrt{(3-2)^2+(3-6)^2}$

$AC=\sqrt{10}$

Checking right angled triangle:

We can check it by Pythagoras theorem

$AB^2=BC^2+AC^2$

now, we can plug values

$(\sqrt{20})^2=(\sqrt{10})^2+(\sqrt{10})^2$

$20=10+10$

$20=20$

we can see that

left side is equal to right side

so, triangle ABC is right angled triangle.........Answer