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Sagot :

KAntoineDoixidn

✒️DISTANCE

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[tex] \large\underline{\mathbb{ANSWER}:} [/tex]

[tex] \qquad \LARGE \quad \rm x = 4 [/tex]

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[tex] \large\underline{\mathbb{SOLUTION}:} [/tex]

Use the distance formula to determine the value of x as x[tex]_2[/tex] with the given distance and points.

[tex] \begin{align} & \bold{Formula:} \\ & \quad \boxed{\rm d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2\,}} \end{align} [/tex]

  • [tex] 10 = \sqrt{\big[x - (\text-2)\big]^2 + (3 - 11)^2\,} [/tex]

  • [tex] 10 = \sqrt{(x + 2)^2 + (\text-8)^2\,} [/tex]

  • [tex] 10 = \sqrt{x^2 + 4x + 4 + 64\,} [/tex]

  • [tex] 10 = \sqrt{x^2 + 4x + 68\,} [/tex]

  • [tex] 10^2 = \big(\sqrt{x^2 + 4x + 68\,}\,\big)^2 [/tex]

  • [tex] 100 = x^2 + 4x + 68\ [/tex]

  • [tex] x^2 + 4x + 68 - 100 = 0 [/tex]

  • [tex] x^2 + 4x - 32 = 0 [/tex]

Solve the quadratic equation by factoring.

  • [tex] (x + 8)(x - 4) = 0 [/tex]

  • [tex] x + 8 = 0 \quad;\quad x - 4 = 0 [/tex]

  • [tex] x = \text-8 \quad;\quad x = 4 [/tex]

Thus, we have two solutions for x. But since it lies on the first quadrant, x must be positive. Therefore, the value of x is 4

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