Math, asked by Taris324, 7 months ago

The points (9,-9),(8,-2),(1,-3)are the vertices of which triangle

Answers

Answered by MaheswariS
3

\underline{\textbf{Given:}}

\textsf{Points are (9,-9), (8,-2), (1,-3)}

\underline{\textbf{To find:}}

\textsf{The type of triangle formed by the given points}

\underline{\textbf{Solution:}}

\textsf{We apply distance formula to the distance}

\textsf{Let the given points be A(9,-9), B(8,-2), C(1,-3)}

\mathsf{AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}

\mathsf{AB=\sqrt{(9-8)^2+(-9+2)^2}}

\mathsf{AB=\sqrt{1^2+(-7)^2}}

\mathsf{AB=\sqrt{1+49}}

\mathsf{AB=\sqrt{50}}

\mathsf{BC=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}

\mathsf{BC=\sqrt{(8-1)^2+(-2+3)^2}}

\mathsf{BC=\sqrt{7^2+1^2}}

\mathsf{BC=\sqrt{49+1}}

\mathsf{BC=\sqrt{50}}

\mathsf{AC=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}

\mathsf{AC=\sqrt{(9-1)^2+(-9+3)^2}}

\mathsf{AC=\sqrt{8^2+(-6)^2}}

\mathsf{AC=\sqrt{64+36}}

\mathsf{AC=\sqrt{100}}

\mathsf{AC=10}

\implies\mathsf{AB=BC\;\;and}

\mathsf{AB^2+BC^2=50+50=100=AC^2}

\therefore\textbf{ABC is an isoceles right angled triangle}

\underline{\textbf{Distance formula:}}

\boxed{\begin{minipage}{8cm}$\\\mathsf{The\;distance\;between\;the\;points(x_1,y_)\;and\;(x_2,y_2)\;is}\\\\\;\;\;\;\mathsf{\;\;\;\;d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}\\$\end{minipage}}

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