Question

Prove that the following points are the vertices of a right triangle.

(-4 ,-3),(-2,2) and (8, -2)


(inappropriate answers will be reported) ​

Answer 1

Answer:

(-4,-3) IS THE RIGHT ANSWER PL FOLLOW ME ON THE BRAINLY

Answer 2

Given,

• The three points of Δ are A(-4,-3) , B(-2,2) and C(8,-2)

We know that,

• The distance b/w two points is given by

$\boxed{ \tt{Distance = \sqrt{ {( x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} } }}$

Thus,

$\tt \pink\implies \purple{AB = \sqrt{ {( - 2 + 4)}^{2} + {(2 + 3)}^{2} }}$

$\tt \blue\implies \green{AB = \sqrt{ {(2)}^{2} + {(5)}^{2} }}$

$\tt\red \implies \orange{AB = \sqrt{4 + 25}}$

$\tt \purple\implies \pink{AB = \sqrt{29}}$

Similarly,

$\tt\green \implies\blue {BC = \sqrt{ {(8 + 2)}^{2} + {( - 2 - 2)}^{2} }}$

$\tt \orange\implies \red{BC = \sqrt{ {(10)}^{2} + {( - 4)}^{2} }}$

$\tt \pink\implies\purple{BC = \sqrt{100 + 16}}$

$\tt \blue\implies \green{BC = \sqrt{116}}$

Now,

$\tt\red \implies \orange{AC = \sqrt{ {(8 + 4)}^{2} + {( - 2 + 3)}^{2} }}$

$\tt \purple\implies \pink{AC = \sqrt{ {(12)}^{2} + {(1)}^{2} }}$

$\tt\green \implies \blue{AC = \sqrt{144 + 1}}$

$\tt \orange\implies\red{ AC = \sqrt{145}}$

It is observed that,

• (AC)² = (AB)² + (BC)²

Therefore , it is a vertices of right angled triangle

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