Question

даны точки A(1;2) B(-3;0) C(4;-2) Найти |3АВ+АС|

Answer 1

A(1;2), B(-3;0), C(4;-2)
AB{-3-1;0-2}, AB{-4;-2}
3*AB{-4*3;-2*3},  3*AB{-12;-6}
AC{4-1;-2-2},  AC{3;-4}
3AB+AC{-12+3;-6+(-4)},   3AB+AC{-9;-10}
|3AB+AC|=√((-9)²+(-10)²)=√(81+100)
|3AB+AC|=√181