How do I prove that $(-3, 0), (0,4), (3,0)$ are the vertices of an isosceles triangle?

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How do I prove that $(-3, 0), (0,4), (3,0)$ are the vertices of an isosceles triangle?

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Answer 1 · 2016-01-05T05:29:52

The sides $P1P2=P2P3=5$ but $P3P1=3$ => $triangle_(P1P2P3)$ is isosceles

Explanation:

Find the size of the sides of the triangle and show that two sides are equal but the third one is different than the others.

$P1P2=sqrt((0+3)^2+(4-0)^2)=sqrt(9+16)=sqrt(25)=5$
$P2P3=sqrt((3-0)^2+(0-4)^2)=sqrt(9+16)=sqrt(25)=5$
$P3P1=sqrt((-3-3)^2+(0-0)^2)-=sqrt(9+0)=sqrt(9)=3$

The sides $P1P2=P2P3=5$ but $P3P1=3$ => $triangle_(P1P2P3)$ is isosceles

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How do I prove that $(-3, 0), (0,4), (3,0)$ are the vertices of an isosceles triangle?

1 Answer

Explanation:

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How do I prove that (-3, 0), (0,4), (3,0)(-3, 0), (0,4), (3,0) are the vertices of an isosceles triangle?

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How do I prove that $(-3, 0), (0,4), (3,0)$ are the vertices of an isosceles triangle?