Question

Without using Pythagoras theorem show that (12,8), (-2,6) and (6,0) are the vertices of right angled triangle.

Answer 1

The vertices of  right angled triangle are A(12, 8), B(- 2, 6) and C(6, 0), proved.

Step-by-step explanation:

Given,

Let the vertices of  right angled triangle are A(12, 8), B(- 2, 6) and C(6, 0).

To prove that,  (12, 8), (- 2, 6) and (6, 0) are the vertices of an right angled triangle.

Using the distance formula,

$\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

AB $= \sqrt{(-2-12)^2+(6-8)^2}$

$= \sqrt{(-14)^2+(-2)^2}$

$= \sqrt{196+4}=\sqrt{200} =10\sqrt{2}$

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

$= \sqrt{(8)^2+(-6)^2}$

$= \sqrt{64+36}=\sqrt{100} =10$

CA$= \sqrt{(12-6)^2+(8-0)^2}$

$= \sqrt{(6)^2+(8)^2}$

$= \sqrt{36+64}$ $= \sqrt{100}$ = 10

Thus, the vertices of  right angled triangle are A(12, 8), B(- 2, 6) and C(6, 0), proved.

Answer 2

Answer:

For without using Pythagoras theorem this attachment will help you:

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