Question

Do the points (3,2),(-2,-3),and (2,3) form a triangle? If so, name type of triangle formed.


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Answer 1

Answer:

Yes, the points ( 3, 2 ), ( - 2, - 3 ) and ( 2, 3 ) form a triangle.

The type of triangle is scalene triangle.

Step-by-step-explanation:

Let the three points be A, B and C.

$\sf\:A\:\equiv\:(\:3,\:2\:)\:=\:(\:x_{1}\:,\:y_{1}\:)\\\\\sf\:B\:\equiv\:(\:-\:2\:,\:-\:3\:)\:\equiv\:(\:x_{2}\:,\:y_{2}\:)\\\\\sf\:C\:\equiv\:(\:2\:,\:3\:)\:\equiv\:(\:x_{3}\:,\:y_{3}\:)$

Now, by using distance formula,

$\pink{\sf\:d\:(\:A,\:B\:)\:=\:\sqrt{\:(\:x_{1}\:-\:x_{2}\:)^{2}\:+\:(\:y_{1}\:-\:y_{2}\:)^{2}\:}}\\\\\implies\sf\:AB\:=\:\sqrt{\:[\:3\:-\:(\:-\:2\:)\:]^{2}\:+\:[\:2\:-\:(\:-\:3\:)\:]^{2}\:}\\\\\implies\sf\:AB\:=\:\sqrt{\:(\:3\:+\:2\:)^{2}\:+\:(\:2\:+\:3\:)^{2}\:}\\\\\implies\sf\:AB\:=\:\sqrt{\:(\:5\:)^{2}\:+\:(\:5\:)^{2}\:}\\\\\implies\sf\:AB\:=\:\sqrt{25\:+\:25\:}\\\\\implies\boxed{\red{\sf\:AB\:=\:\sqrt{50}}}\:\:\:\sf\:-\:-\:(\:1\:)$

Now,

$\sf\:BC\:=\:\sqrt{\:(\:-\:2\:-\:2\:)^{2}\:+\:(\:-\:3\:-\:3\:)^{2}\:}\\\\\implies\sf\:BC\:=\:\sqrt{\:(\:-\:4\:)^{2}\:+\:(\:-\:6\:)^{2}\:}\\\\\implies\sf\:BC\:=\:\sqrt{\:16\:+\:36\:}\\\\\implies\boxed{\pink{\sf\:BC\:=\:\sqrt{52}}}\:\sf\:\:-\:-\:(\:2\:)$

Now,

$\sf\:AC\:=\:\sqrt{\:(\:3\:-\:2\:)^{2}\:+\:(\:2\:-\:3\:)^{2}\:}\\\\\implies\sf\:AC\:=\:\sqrt{\:(\:1\:)^{2}\:+\:(\:-\:1\:)^{2}\:}\\\\\implies\sf\:AC\:=\:\sqrt{\:1\:+\:1\:}\\\\\implies\boxed{\pink{\sf\:AC\:=\:\sqrt{2}}}\:\sf\:\:-\:-\:-\:(\:3\:)$

From ( 1 ), ( 2 ) & ( 3 ),

BC is the greatest segment.

We know that,

To form a triangle, the sum of two sides must be greater than the third side.

$\therefore\sf\:AB\:+\:AC\:=\:\sqrt{50}\:+\:\sqrt{2}$

We have to verify that

$\sf\:\sqrt{50}\:+\:\sqrt{2}\:>\:\sqrt{52}$

Now,

$\sf\:(\:\sqrt{50}\:+\:\sqrt{2}\:)^{2}\\\\\implies\sf\:(\:\sqrt{50}\:)^{2}\:+\:2\:\times\:\sqrt{50}\:\times\:\sqrt{2}\:+\:(\:\sqrt{2}\:)^{2}\:\:\:-\:-\:[\:(\:a\:+\:b\:)^{2}\:=\:a^{2}\:+\:2ab\:+\:b^{2}\:]\\\\\implies\sf\:50\:+\:2\:\times\:\sqrt{100}\:+\:2\\\\\implies\sf\:50\:+\:2\:\times\:10\:+\:2\\\\\implies\sf\:50\:+\:20\:+\:2\\\\\implies\sf\:70\:+\:2\\\\\implies\boxed{\red{\sf\:72}} \:\:\:-\:\sf\:-\:-\:(\:4\:)\\\\\sf\:(\:\sqrt{52}\:)^{2}\\\\\implies\sf\:\sqrt{52}\:\times\:\sqrt{52}\\\\\implies\boxed{\red{\sf\:52}}\:\:\sf\:\:-\:-\:(\:5\:)$

Now,

$\sf\:72\:>\:52\\\\\implies\sf\:(\:\sqrt{50}\:+\:\sqrt{2}\:)^{2}\:>\:(\:\sqrt{52}\:)^{2}\:\:\:-\:-\:[\:From\:(\:4\:)\:\&\:(\:5\:)\:]\\\\\implies\sf\:\sqrt{50}\:+\:\sqrt{2}\:>\:\sqrt{52}\:\:\:-\:-\:-\:[\:Taking\:square\:roots\:]\\\\\therefore\sf\:\triangle\:ABC\:is\:formed\\\\\sf\:AB\:\neq\:BC\:\neq\:AC\\\\\therefore\sf\:\triangle\:ABC\:is\:scalene\:triangle.$