# How do you find x so distance between the points (6, -1) and (x, 9) is 12?

Jun 12, 2016

Problems like this require the involvement of the distance formula, $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

#### Explanation:

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$12 = \sqrt{{\left(x - 6\right)}^{2} + {\left(9 - \left(- 1\right)\right)}^{2}}$

$12 = \sqrt{{x}^{2} - 12 x + 36 + 100}$

12 = sqrt(x^2 - 12x + 136

${\left(12\right)}^{2} = {\left(\sqrt{{x}^{2} - 12 x + 136}\right)}^{2}$

$144 = {x}^{2} - 12 x + 136$

$0 = {x}^{2} - 12 x - 8$

By the completion of square method:

$0 = 1 \left({x}^{2} - 12 x + n\right) - 8$

$n = {\left(\frac{b}{2}\right)}^{2}$

$n = {\left(- \frac{12}{2}\right)}^{2}$

$n = 36$

$0 = 1 \left({x}^{2} - 12 x + 36 - 36\right) - 8$

$0 = 1 \left({x}^{2} - 12 x + 36\right) - 36 - 8$

$0 = 1 {\left(x - 6\right)}^{2} - 44$

$44 = {\left(x - 6\right)}^{2}$

$\pm \sqrt{44} = x - 6$

$\pm \sqrt{44} + 6 = x$

$\pm 2 \sqrt{11} + 6 = x$

$x = 2 \left(3 - \sqrt{11}\right) \mathmr{and} 2 \left(3 + \sqrt{11}\right)$

Checking back in the original equation, you will find both solutions work.

Hopefully this helps!