How to find the sum of the squares of the lengths of the legs of a triangle of a=(-6,11) b=(2,8) c=(-1,0)?

We have that
a=(-6,11) b=(2,8) c=(-1,0)

step 1
find the distances with the formula
d=√[(y2-y1)²+(x2-x1)²]

distance ab
dab=√[(8-11)²+(2+6)²]> dab=√73 units

distance bc
dbc=√[(0-8)²+(-1-2)²]> bc=√73 units

distance ac
dac=√[(0-11)²+(-1+6)²]> dac=√146 units

the legs are 
ab=√73 units
bc=√73 units

the hypotenuse is 
ac=√146 units

the sum of the squares of the lengths of the legs is
(√73)²+(√73)²=73+73> 146

the answer is
146

B-simple yes/no questions

The total would be 40.53