Question

AABC has vertices at A(1, -9), B(8,0), and C(9,-8).

Is AABC an equilateral triangle? Justify your answer.

Answer 1

Check the picture below.

`~\hfill \stackrel{\textit{\large distance between 2 points}}{d = √(( 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=√([ 8- 1]^2 + [ 0- (-9)]^2) \\\\\\ AB=√(7^2+(0+9)^2)\implies AB=√(7^2+9^2)\implies \boxed{AB=√(130)} \\\\[-0.35em] ~\dotfill`

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

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