If the distance between the points $(x, 2)$ and $(3, - 6)$ is $10$ units, what is the positive value of $x$ .

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Solution

Distance between two points
$\sqrt{{(x 2 - x 1)}^{2} + {(y 2 - y 1)}^{2}}$
$\sqrt{{(x - 3)}^{2} + {(2 - (- 6))}^{2}} = 10$
$\sqrt{{(x - 3)}^{2} + 64} = 10$
Squaring both sides
$(x - 3)^{2} + 64 = 100$
$x^{2} - 6 x - 27 = 0$
$x^{2} - 9 x + 3 x - 27 = 0$
$x (x - 9) + 3 (x - 9) = 0$
$x = - 3, 9$
Taking positive value, $x = 9 u n i t s$

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