Question

calcule a distancia entre os pontos :
a) A(-1,-1), B(1,1)
b) A(2,3),B (2,5)

Answer 1

a)
d²=(xb-xa)²+(yb-ya)²
d²=(1+1)²+(1+1)²
d²=(2)²+(2)²
d²=4+4
d²=8
d=raiz de 8

b)
d²=(xb-xa)²+(yb-ya)²
d²=(2-2)²+(5-3)²
d²=2²
d²=4
d=raiz de 4
d=2

Answer 2

a)

$\large \boxed{ \begin{array}{l} \sf{d_{AB} = \sqrt{(x_A - x_B) {}^{2} + (y_A - y_B) {}^{2} } } \\ \\ \sf{d_{AB} = \sqrt{ ((- 1) - 1) {}^{2} + (( - 1) - 1) {}^{2} } } \\ \\ \sf{d_{AB} = \sqrt{( - 2) {}^{2} + ( - 2) {}^{2} } } \\ \\ \sf{d_{AB} = \sqrt{4 + 4} } \\ \\ \sf{d_{AB} = \sqrt{8} } \\ \\ \boxed{ \boxed{ \sf{d_{AB} = 2 \sqrt{2} }}}\end{array}}$

b)

$\large \boxed{ \begin{array}{l} \sf{d_{AB} = \sqrt{(x_A - x_B) {}^{2} + (y_A - y_B) {}^{2} } } \\ \\ \sf{d_{AB} = \sqrt{(2 - 2) {}^{2} + (3 - 5) {}^{2} } } \\ \\ \sf{d_{AB} = \sqrt{0 {}^{2} + ( - 2) {}^{2} } } \\ \\ \sf{d_{AB} = \sqrt{0 + 4} } \\ \\ \sf{d_{AB} = \sqrt{4} } \\ \\ \boxed{ \boxed{ \sf{d_{AB} = 2}}}\end{array}}$