Answered

Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

AABC has vertices at A(1, -9), B(8,0), and C(9,-8).
Is AABC an equilateral triangle? Justify your answer.

Sagot :

Check the picture below.

[tex]~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{-9})\qquad B(\stackrel{x_2}{8}~,~\stackrel{y_2}{0}) ~\hfill AB=\sqrt{[ 8- 1]^2 + [ 0- (-9)]^2} \\\\\\ AB=\sqrt{7^2+(0+9)^2}\implies AB=\sqrt{7^2+9^2}\implies \boxed{AB=\sqrt{130}} \\\\[-0.35em] ~\dotfill[/tex]

[tex]B(\stackrel{x_1}{8}~,~\stackrel{y_1}{0})\qquad C(\stackrel{x_2}{9}~,~\stackrel{y_2}{-8}) ~\hfill BC=\sqrt{[ 9- 8]^2 + [ -8- 0]^2} \\\\\\ BC=\sqrt{1^2+(-8)^2}\implies \boxed{BC=\sqrt{65}}[/tex]

now, we could check for the CA distance, however, we already know that AB ≠ BC, so there's no need.

View image jdoe0001
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.