Question

Triangle abc has vertices located at a( 0, 2), b (2,5), and c (-1,7). find the length of each side of the triangle

Answer 1

$ab=\sqrt{(0-2)^{2}+(2-5)^{2}}=\boxed{\sqrt{13}}\\\\bc=\sqrt{(2-(-1))^{2}+(5-7)^{2}}=\boxed{\sqrt{13}}\\\\ac=\sqrt{(0-(-1))^{2}+(2-7)^{2}}=\boxed{\sqrt{26}}$