Answer:
Let \[A(3,0),\text{ }B(6,4)\] and \[C(-1,3)\] be the vertices of a triangle \[ABC\]. Length of \[AB=\sqrt{{{(6-3)}^{2}}+{{(4-0)}^{2}}}\] \[=\sqrt{{{(3)}^{2}}+{{(4)}^{2}}}\] \[=\sqrt{9+16}=\sqrt{25}=5\,\]units Length of \[BC=\sqrt{{{(-1-6)}^{2}}+{{(3-4)}^{2}}}\] \[=\sqrt{{{(-7)}^{2}}+{{(-1)}^{2}}}\] \[=\sqrt{49+1}=\sqrt{50}=5\sqrt{2}\] units. And Length of \[AC=\sqrt{{{(-1-3)}^{2}}+{{(3-0)}^{2}}}\] \[=\sqrt{{{(-4)}^{2}}+{{(3)}^{2}}}\] \[=\sqrt{16+9}=\sqrt{25}=5\] units \[\therefore \] \[AB=AC\] And \[{{(AB)}^{2}}+{{(AC)}^{2}}={{(BC)}^{2}}\] Hence, \[\Delta \text{ }ABC\] is a isosceles, right angled triangle. Hence Proved
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