Respuesta :

dalendrk
[tex]A(x_A;\ y_A);\ B(x_B;\ y_B).\\\\Distance\ d\ between\ A\ and\ B:\\\\d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}[/tex]


[tex]P(-4;\ 3);\ Q(6;\ 1)\\\\|PQ|=\sqrt{(6-(-4))^2+(1-3)^2}=\sqrt{10^2+(-2)^2}=\sqrt{100+4}\\\\=\sqrt{104}=\sqrt{4\cdot26}=\sqrt4\cdot\sqrt{26}=2\sqrt{26}[/tex]
Amphitrite1040
The distance  between P ( -4, 3)  & Q (6,1)

[tex]PQ = \sqrt{104} = 10.2[/tex]

So, 10.2 is your answer

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