Question

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

Answer 1

$\underline{\textbf{Given:}}$

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

$\underline{\textbf{To find:}}$

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

$\underline{\textbf{Solution:}}$

$\textsf{We apply distance formula to the distance}$

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

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

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

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

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

$\mathsf{AB=\sqrt{50}}$

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

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

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

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

$\mathsf{BC=\sqrt{50}}$

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

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

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

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

$\mathsf{AC=\sqrt{100}}$

$\mathsf{AC=10}$

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

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

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

$\underline{\textbf{Distance formula:}}$

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