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Question
Are the points A(3, 6, 9), B(10, 20, 30) and C(25, –41, 5), the vertices of a right-angled triangle?
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Solution
Let A(3,6,9), B(10,20,30) and C( 25,\[-\]41,5) are vertices of \[\bigtriangleup ABC\]
AB =\[\sqrt{\left( 10 - 3 \right)^2 + \left( 20 - 6 \right)^2 + \left( 30 - 9 \right)^2}\]
\[= \sqrt{\left( 7 \right)^2 + \left( 14 \right)^2 + \left( 21 \right)^2}\]
\[ = \sqrt{49 + 196 + 441}\]
\[ = \sqrt{686}\]
\[ = 7\sqrt{14}\]
BC =\[\sqrt{\left( 25 - 10 \right)^2 + \left( - 41 - 20 \right)^2 + \left( 5 - 30 \right)^2}\]
\[= \sqrt{\left( 15 \right)^2 + \left( - 61 \right)^2 + \left( - 25 \right)^2}\]
\[ = \sqrt{225 + 3721 + 625}\]
\[ = \sqrt{4571}\]
CA=\[\sqrt{\left( 3 - 25 \right)^2 + \left( 6 + 41 \right)^2 + \left( 9 - 5 \right)^2}\]
\[= \sqrt{\left( - 22 \right)^2 + \left( 47 \right)^2 + \left( - 4 \right)^2}\]
\[ = \sqrt{484 + 2209 + 16}\]
\[ = \sqrt{2709}\]
\[ = 3\sqrt[]{301}\]
\[A B^2 + B C^2 = \left( 7\sqrt{14} \right)^2 + \left( \sqrt{4571} \right)^2 \]
\[ = 686 + 4571\]
\[ = 5257\]
\[C A^2 = 2709\]
\[ \therefore A B^2 + B C^2 \neq C A^2\]
A triangle\[\bigtriangleup ABC\]is right-angled at B if \[C A^2 = A B^2 + B C^2\]
But,\[C A^2\] ≠\[A B^2 + B C^2\]Hence, the points are not vertices of a right-angled triangle.
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