Question

Prove that by joining the points (2,1) (3,4) (-3,6) we get a right triangle​

Answer 1

You could verify this by plotting points to graph, (even without marking points), just approximately you can verify.

We know that,

• $\sf{A\:(x_1,\:y_1)\:=\:(2,\:1)}$
• $\sf{A\:(x_2,\:y_2)\:=\:(3,\:4)}$

$\implies\sf{AB\:=\:{\sqrt{(x_2\:-\:x_1)^{2}\:+\:(y_2\:-\:y_1)^{2}}}}$

$\implies\sf{AB\:=\:{\sqrt{(3\:-\:2)^{2}\:+\:(4\:-\:1)^{2}}}}$

$\implies\sf{AB\:=\:{\sqrt{(1)^{2}\:+\:(3)^{2}}}}$

$\implies\sf{AB\:=\:{\sqrt{1\:+\:9}}}$

$\implies\sf{AB\:=\:{\sqrt{10}}}$

Secondly,

• $\sf{B\:(x_1,\:y_1)\:=\:(3,\:4)}$
• $\sf{C\:(x_2,\:y_2)\:=\:(-3,\:6)}$

$\implies\sf{BC\:=\:{\sqrt{(x_2\:-\:x_1)^{2}\:+\:(y_2\:-\:y_1)^{2}}}}$

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

$\implies\sf{BC\:=\:{\sqrt{(6)^{2}\:+\:(2)^{2}}}}$

$\implies\sf{BC\:=\:{\sqrt{36\:+\:4}}}$

$\implies\sf{BC\:=\:{\sqrt{40}}}$

Lastly,

• $\sf{C\:(x_1,\:y_1)\:=\:(-3,\:6)}$
• $\sf{CA\:(x_2,\:y_2)\:=\:(2,\:1)}$

$\implies\sf{A\:=\:{\sqrt{(x_2\:-\:x_1)^{2}\:+\:(y_2\:-\:y_1)^{2}}}}$

$\implies\sf{CA\:=\:{\sqrt{(2\:-\:(-3))^{2}\:+\:(1\:-\:6)^{2}}}}$

$\implies\sf{CA\:=\:{\sqrt{(5)^{2}\:+\:(-5)^{2}}}}$

$\implies\sf{CA\:=\:{\sqrt{25\:+\:25}}}$

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

By using Pythagoras Theorem,

$\boxed{AC^2\:=\:AB^2\:+\:BC^2}$

Substituting values, we get,

$\implies\sf{{\sqrt{50}}^2\:=\:{\sqrt{10}}^2\:+\:{\sqrt{40}}^2}$

$\implies\sf{50\:=\:10\:+\:40}$

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

Hence, Joining the points (2,1), (3,4) and (-3,6) we get a right angled triangle.

Answer 2

Answer:

Let, A(2, 1), B(3, 4), C(-3, 6) be the vertices of the triangle ABC.

$ab \: = \sqrt{ {(3 - 2)}^{2} + {(4 - 1)}^{2} }$

$= \sqrt{ {1}^{2} + {3}^{2} } = \sqrt{10}$

$bc = \sqrt{ { (- 3 - 3)}^{2} + {(6 - 4)}^{2} }$

$= \sqrt{ {6}^{2} + {2}^{2} } = \sqrt{40}$

$ac \: = \sqrt{ {( 2 - 3)}^{2} \: + {(1 - 6)}^{2} }$

$\sqrt{ {5}^{2} + {( - 5)}^{2} } = \sqrt{50}$

Using pythagoras theorem

${ac}^{2} = {ab}^{2} + {bc}^{2}$

$50 = {(√10)}^{2} + {( \sqrt{40} }^{2} = 50$

Therefore, we obtain a right angle triangle