How do you show that a triangle with vertices $(13,-2)$ , $(9,-8)$ , $(5,-2)$ is isosceles?

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How do you show that a triangle with vertices $(13,-2)$ , $(9,-8)$ , $(5,-2)$ is isosceles?

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Answer 1 · 2016-01-06T23:59:19

Find the length of the triangle's segments and prove that two are equal but the third is different. In case, $P1P1=P2P3=sqrt(50)$ and $P1P3=8$ => $triangle_(P1P2P3)$ is isosceles

Explanation:

Length of the triangle's segments:
$P1P2=sqrt((9-13)^2+(-8+2)^2)=sqrt (16+36)=sqrt(50)$
$P1P3=sqrt((5-13^2+(-2+2)^2)=sqrt(64+0)=8$
$P2P3=sqrt((5-9)^2+(-2+8)^2)=sqrt(16+36)=sqrt(50$

As we can see two sides of the triangle are equal ( $P1P2=P2P3=sqrt(50)$ ) but the third one is different ( $P1P3=8$ ): so $triangle_(P1P2P3)$ is isosceles.

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How do you show that a triangle with vertices $(13,-2)$ , $(9,-8)$ , $(5,-2)$ is isosceles?

1 Answer

Explanation:

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How do you show that a triangle with vertices (13,-2)(13,-2), (9,-8)(9,-8), (5,-2)(5,-2) is isosceles?

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How do you show that a triangle with vertices $(13,-2)$ , $(9,-8)$ , $(5,-2)$ is isosceles?