The sum of the squares of the lengths of the legs of the triangle when A=(-8,11) B=(1,9) C=(-1,0)

Question

The sum of the squares of the lengths of the legs of the triangle

Answer 1

Slope of AB=(9-11)/(1-(-8))=-2/9; the slope of BC=(9-0)/(1-(-1))=9/2. These slopes are perpendicular to one another because -2/9 x 9/2 = -1.

Therefore the angle B is 90 degrees. The side lengths are BC=AB=√(9^2+2^2)=√85; AC=√(11^2+7^2)=√170. (Using Pythagoras and counting squares or working out the difference between coords.)

AC^2=170=AB^2+BC^2=85+85=170 in accordance with Pythagoras and the triangle is right isosceles.