Question

Prove that the points (3,0), (6,4) and (-1,3) are the vertices of a right angled isosceles triangle.

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Answer 1

Since AB2+AC2=BC2 and AB=AC

∴ ABC is a right angled isosceles triangle

Answer 2

Question :-

Prove that the points (3,0), (6,4) and (-1,3) are the vertices of a right angled isosceles triangle.

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Solution :-

Let A(3, 0), B(6,4) and C(-1,3) be the

given points.

$\red{\maltese}\:\sf{ab = \sqrt{(6 - 3)^{2} + {(4 - 0)}^{2} }}$

$\sf: \implies AB= \sqrt{ {(3)}^{2} + {(4)}^{2} }$

$\sf: \implies AB = \sqrt{9 + 16}$

$\sf: \implies AB = \sqrt{25}$

$\red{ \maltese} \: \sf {BC = \sqrt{ {( - 1 - 6)}^{2} + (3 - 4)^{2} }}$

$\sf : \implies BC = \sqrt{ {( - 7)}^{2} + ( - 1)^{2} }$

$\sf: \implies BC = \sqrt{49 + 1}$

$\sf : \implies BC = \sqrt{50}$

$\red{ \maltese} \: \sf{AC = \sqrt{ {( - 1 - 3)}^{2} + (3 - 0)^{2} }}$

$\sf: \implies AC = \sqrt{ {( - 4)}^{2} + {(3)}^{2} }$

$\sf : \implies AC = \sqrt{16 + 9}$

$\sf : \implies AC = \sqrt{25}$

$\sf : \implies {AB}^{2} = {({ \sqrt{25} })}^{2}$

$\sf : \implies {AB}^{2} = 25$

$\sf : \implies {AC}^{2} = 25$

$\sf: \implies {BC}^{2} = {( \sqrt{50} )}^{2}$

$\sf : \implies {BC}^{2} = 50$

$\sf \because {AB}^{2} + {AC}^{2} = {BC}^{2}$

$\sf \: and \: AB = AC$

$\sf \therefore \triangle \: ABC \: is \: a\: right \: angled \: isosceles \: triangle$

Hence Proved

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