Question

calcule a distância de dois pontos: ​

Answer 1

Resposta:

a) $d_{AB} =\sqrt{41}$      b) $d_{PQ} =5\sqrt{2}$      c) $d_{RS} =5,5$    

d) $d_{TV} =\sqrt{113}$     e) $d_{AB} =8$         f ) $d_{AB} =4\sqrt{5}$    

g) $d_{AB} =5\sqrt{2}$      h) $d_{AB} =1$         i ) $d_{AB} =5$

Explicação passo a passo:

Lembrete: Caso for uma reta, ou seja, se um dos 2 valores referênciais (linha (x) OU coluna (y)) forem iguais, então será a diferença do outro. (Caso das alternativas: 'c', 'e' e 'h').

a) A (2, 5) e B (-3, 1)

$d_{AB} =\sqrt{(-3-2)^{2}+(1-5)^{2}} \\\\d_{AB} =\sqrt{(-5)^{2}+(-4)^{2}} \\\\d_{AB} =\sqrt{25+16} \\\\\bold{d_{AB} =\sqrt{41}}$

b) P (-4, 0) e Q (1, -5)

$d_{PQ} =\sqrt{(1-(-4))^{2}+(-5-0)^{2}} \\\\d_{PQ} =\sqrt{(5)^{2}+(-5)^{2}} \\\\d_{PQ} =\sqrt{25+25} \\\\d_{PQ} =\sqrt{50} \\\\d_{PQ} =\sqrt{25\times2} \\\\d_{PQ} =\sqrt{25}\sqrt{2} \\\\d_{PQ} =\sqrt{5^{2} } \sqrt{2} \\\\\bold{d_{PQ} =5\sqrt{2} }$

c) R (-1/2, 1) e S (5, 1)

$d_{RS} =5-(-\frac{1}{2} )\\\\d_{RS} =5+\frac{1}{2} \\\\d_{RS} =5+0,5 \\\\\bold{d_{RS} =5,5 }$

d) T (0, -8) e V (-7, 0)

$d_{TV} =\sqrt{(-7-0)^{2}+(0-(-8))^{2}} \\\\d_{TV} =\sqrt{(-7)^{2}+(8)^{2}} \\\\d_{TV} =\sqrt{49+64} \\\\\bold{d_{TV} =\sqrt{113}}$

e) A (-3, 4) e B (5, 4)

$d_{AB}= 5-(-3)\\\\d_{AB}= 5+3\\\\\bold{d_{AB}= 8}$

f) A (-2, 5) e B (6, 1)

$d_{AB} =\sqrt{(6-(-2))^{2}+(1-5)^{2}} \\\\d_{AB} =\sqrt{8^{2}+(-4)^{2}} \\\\d_{AB} =\sqrt{64+16} \\\\d_{AB} =\sqrt{80} \\\\d_{AB} =\sqrt{16\times5} \\\\d_{AB} =\sqrt{16}\sqrt{5} \\\\d_{AB} =\sqrt{4^{2} }\sqrt{5} \\\\\bold{d_{AB} =4\sqrt{5}}$

g) A (-2, 3) e B (5, 2)

$d_{AB} =\sqrt{(5-(-2))^{2}+(2-3)^{2}} \\\\d_{AB} =\sqrt{7^{2}+(-1)^{2}} \\\\d_{AB} =\sqrt{49+1} \\\\d_{AB} =\sqrt{50} \\\\d_{AB} =\sqrt{25\times2} \\\\d_{AB} =\sqrt{25}\sqrt{2} \\\\d_{AB} =\sqrt{5^{2} }\sqrt{2} \\\\\bold{d_{AB} =5\sqrt{2}}$

h) A (1, 1) e B (2, 1)

$d_{AB} =2-1\\\\\bold{d_{AB} =1}$

i) A (1, -2) e B (-3, 1)

$d_{AB} =\sqrt{(-3-1)^{2}+(1-(-2))^{2}} \\\\d_{AB} =\sqrt{(-4)^{2}+3^{2}} \\\\d_{AB} =\sqrt{16+9} \\\\d_{AB} =\sqrt{25} \\\\d_{AB} =\sqrt{5^{2} } \\\\\bold{d_{AB} =5}$