Question

3) Sabendo que a distância entre os pontos A (-2, y) e B (6, 7) é 10. Qual o valor de y?​

Answer 1

<var>d=

(x

f

−x

i

)

2

+(y

f

−y

i

)

2

</var>

\begin{gathered} < var > 10 = \sqrt{(6 - (-2))^{2} + (7-y)^{2}} \\\\ 10 = \sqrt{(6+2)^{2} + (7-y)^{2}} \\\\ 10 = \sqrt{(8)^{2} + (7-y)^{2}} \\\\ 10 = \sqrt{64 + (7-y)^{2}} \\\\ elevamos \ os \ dois \ lados \ ao \ quadrado \ para \ sumir \ com \ a \ raiz \\\\ (10)^{2} = (\sqrt{64 + (7-y)^{2}})^{2} \\\\ 100 = 64 + (7-y)^{2} \\\\ 100 = 64 + 49 - 14y + y^{2} \\\\ y^{2} - 14y + 49 + 64 - 100 = 0 \\\\ y^{2} - 14y + 13 = 0 < /var > \end{gathered}

<var>10=

(6−(−2))

2

+(7−y)

2

10=

(6+2)

2

+(7−y)

2

10=

(8)

2

+(7−y)

2

10=

64+(7−y)

2

elevamos os dois lados ao quadrado para sumir com a raiz

(10)

2

=(

64+(7−y)

2

)

2

100=64+(7−y)

2

100=64+49−14y+y

2

y

2

−14y+49+64−100=0

y

2

−14y+13=0</var>

\begin{gathered} < var > \Delta = b^{2} - 4 \cdot a \cdot c \\ \Delta = (14)^{2} - 4 \cdot (1) \cdot (13) \\ \Delta = 196 - 52 \\ \Delta = 144 < /var > \end{gathered}

<var>Δ=b

2

−4⋅a⋅c

Δ=(14)

2

−4⋅(1)⋅(13)

Δ=196−52

Δ=144</var>

\begin{gathered} < var > y = \frac{-b \pm\sqrt{\Delta} }{2 \cdot a} \\\\ y = \frac{-(-14) \pm\sqrt{144} }{2 \cdot 1} \\\\ y = \frac{14 \pm 12}{2} \\\\\\ \rightarrow y' = \frac{14 + 12}{2} \\\\ y' = \frac{26}{2} \\\\ \boxed{y' = 13} \\\\\\ \rightarrow y'' = \frac{14 - 12}{2} \\\\ y'' = \frac{2}{2} \\\\ \boxed{y'' = 1} < /var > \end{gathered}

<var>y=

2⋅a

−b±

Δ

y=

2⋅1

−(−14)±

144

y=

2

14±12

→y

′

=

2

14+12

y

′

=

2

26

y

′

=13

→y

′′

=

2

14−12

y

′′

=

2

2

y

′′

=1

</var>

\begin{gathered} < var > \therefore y \ pode \ ter \ dois \ valores: \\ \rightarrow \boxed{y = 1} \\ \rightarrow \boxed{y = 13} < /var > \end{gathered}

<var>∴y pode ter dois valores:

→

y=1

→

y=13

</var>

Portanto o ponto pode ser (-2, 1) e (-2, 13)