Question

4. Se os pontos A(k − 1, 3) e B(2, −1) sao tais que dAB = 5, entao qual e a abscissa do ponto A?

Answer 1

Resposta:

5

Explicação passo-a-passo:

(k - 1 - 2)² + (3 + 1)² = 5²

(k - 3)² + 4² = 25

(k - 3)² = 25 - 16

(k - 3)² = 9

(k - 3)² = 3²

k - 3 = 3

k = 3 + 3

k = 6

A(k - 1 , 3)

A(6 - 1, 3)

A(5, 3)

Answer 2

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$\large\boxed{\begin{array}{l}\sf Dados~A(x_A,y_A),B(x_B,y_B)~dois~pontos~quaisquer\\\sf do~plano~cartesiano,a~dist\hat ancia~entre~esses\\\sf pontos~\acute e~dada~por\\\sf d_{A,B}=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\end{array}}$

$\small\boxed{\begin{array}{l}\underline{\rm dados\!:}\\\sf d_{A,B}=5\\\sf x_{A}=?\\\sf y_A=3\\\sf x_B=2\\\sf y_B=-1\\\underline{\bf soluc_{\!\!,}\tilde ao\!:}\\\sf x_B-x_A=2-[k-1]=2-k+1=3-k\\\sf y_B-y_A=-1-3=-4\\\sf d_{A,B}=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\\\sf 5=\sqrt{(3-k)^2+(-4)^2}\\\sf\sqrt{(3-k)^2+16}=5\\\sf(\sqrt{(3-k)^2+16)^2}=5^2\\\sf (3-k)^2+16=25\\\sf (3-k)^2=25-16\\\sf (3-k)^2=9\\\sf 3-k=\pm\sqrt{9}\\\sf 3-k=\pm3\\\sf 3-k=3\\\sf k=3-3\\\sf k=0\\\sf 3-k=-3\\\sf k=3+3\\\sf k=6 \end{array}}$

$\boxed{\begin{array}{l}\sf se~k=0:\\\sf x_A=0-1=-1\\\sf se~k=6:\\\sf x_A=6-1=5\end{array}}$