Find the value of x so the distance between the points A(3,5) and B(x,8) is 5 units.

Sunday, March 23, 2014

Find the value of x so the distance between the points A(3,5) and B(x,8) is 5 units.

The problem provides the length of the segment $\displaystyle{A}{B}={5}$ , hence,
using Pythagorean theorem yields:

$\displaystyle{A}{{B}}^{{2}}={{\left({3}-{x}\right)}}^{{2}}+{{\left({5}-{8}\right)}}^{{2}}$

$\displaystyle{{5}}^{{2}}={9}-{6}{x}+{{x}}^{{2}}+{9}\Rightarrow{25}-{18}={{x}}^{{2}}-$
6x

$\displaystyle{{x}}^{{2}}-{6}{x}-{7}={0}$

You may use the
factorization to evaluate the solutions to quadratic equation, such
that:

$\displaystyle{{x}}^{{2}}-{7}{x}+{x}-{7}={0}\Rightarrow{\left({{x}}^{{2}}-{7}{x}\right)}+{\left({x}-{7}\right)}=$
0

$\displaystyle{x}{\left({x}-{7}\right)}+{\left({x}-{7}\right)}={0}\Rightarrow{\left({x}-{7}\right)}{\left({x}+{1}\right)}={0}\Rightarrow$
{(x - 7 = 0),(x + 1 = 0):} => {(x = 7),(x = -1):}

Hence, evaluating the missing coordinate x yields
that there exists two points whose y coordinate is $\displaystyle{y}={8}$ , such that: $\displaystyle{x}={7},{B}{\left({7},{8}\right)}$ and
$\displaystyle{x}=-{1},{B}{\left(-{1},{8}\right)}.$

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Sunday, March 23, 2014