Question

A distância entre os pontos A ( -2, y) e B (6,7) é 10. O valor de y é: *
1 ponto
a) -1
b) 0
c) 1 ou 13
d) -1 ou 10
e) 2 ou 12

Answer 1

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

$\boxed{\begin{array}{l}\sf A(-2,y)~~B(6,7)\\\sf d_{A,B}=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\\\sf 10=\sqrt{(6-[-2])^2+(7-y)^2}\\\sf 10=\sqrt{(6+2)^2+(7-y)^2}\\\sf 10=\sqrt{64+49-14y+y^2}\\\sf\sqrt{y^2-14y+113}=10\\\sf (\sqrt{y^2-14y+113})^2=10^2\\\sf y^2-14y+113=100\\\sf y^2-14y+113-100=0\end{array}}$

$\large\boxed{\begin{array}{l}\sf y^2-14y+13=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-14)^2-4\cdot1\cdot13\\\sf\Delta=196-52\\\sf\Delta=144\\\sf y=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf y=\dfrac{-(-14)\pm\sqrt{144}}{2\cdot1}\\\\\sf y=\dfrac{14\pm12}{2}\begin{cases}\sf y_1=\dfrac{14+12}{2}=\dfrac{26}{2}=13\\\\\sf y_2=\dfrac{14-12}{2}=\dfrac{2}{2}=1\end{cases}\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~c}}}}\end{array}}$