question_answer

Prove that the points (3, 0), (6, 4) and (\[1,\text{ }3\]) are the vertices of a right angled isosceles triangle.

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