Question

A distância do ponto d(AB) é 5. Sendo A(-1,-2), B (3,-y), calcule o valor de y​

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{d_{AB} = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2}}$

$\mathsf{5 = \sqrt{(-1 - 3)^2 + (-2 - (-y))^2}}$

$\mathsf{25 = (-1 - 3)^2 + (-2 - (-y))^2}$

$\mathsf{25 = (-1 - 3)^2 + (y - 2)^2}$

$\mathsf{25 = 16 + (y^2 - 4y + 4)}$

$\mathsf{y^2 - 4y - 5 + 9 = 0 + 9}$

$\mathsf{y^2 - 4y + 4 = 9}$

$\mathsf{(y - 2)^2 = 9}$

$\mathsf{y' - 2 = 3}$

$\mathsf{y' = 5}$

$\mathsf{y'' - 2 = -3}$

$\mathsf{y'' = -1}$

$\boxed{\boxed{\mathsf{S = \{5;-1\}}}}$

Answer 2

Resposta:

$\Large\mathsf{d_{AB} = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2}} \\\Large\mathsf{5 = \sqrt{(-1 - 3)^2 + (-2 - (-y))^2}} \\\Large \mathsf{25 = (-1 - 3)^2 + (-2 - (-y))^2} \\\Large \mathsf{25 = (-1 - 3)^2 + (y - 2)^2} \\ \Large\mathsf{25 = 16 + (y^2 - 4y + 4)} \\ \Large\mathsf{y^2 - 4y - 5 + 9 = 0 + 9} \\\Large\mathsf{y^2 - 4y + 4 = 9} \\\Large \mathsf{(y - 2)^2 = 9} \\ \Large\mathsf{y' - 2 = 3} \\ \Large\mathsf{y' = 5} \\ \Large\mathsf{y'' - 2 = -3} \\ \Large\mathsf{y'' = -1} \\\Large \red{\mathsf{S = \{5;-1\}}}$

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