The distance between P(x, 7) and Q(6, 19) is 13 units. Find the values of x.
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Correct Answer: Option C
Explanation:
\(d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\)
\(13 = \sqrt{(x - 6)^{2} + (7 - 19)^{2}}\)
\(13^{2} = x^{2} - 12x + 36 + 144\)
\(169 = x^{2} - 12x + 180\)
\(x^{2} - 12x + 180 - 169 = 0 \implies x^{2} - 12x + 11 = 0\)
\((x - 1)(x - 11) = 0 \implies x = \text{1 or 11}\)
\(d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\)
\(13 = \sqrt{(x - 6)^{2} + (7 - 19)^{2}}\)
\(13^{2} = x^{2} - 12x + 36 + 144\)
\(169 = x^{2} - 12x + 180\)
\(x^{2} - 12x + 180 - 169 = 0 \implies x^{2} - 12x + 11 = 0\)
\((x - 1)(x - 11) = 0 \implies x = \text{1 or 11}\)