The Notes: `(6, 2), -3x + 4y = -5` Find the distance between the point and the line.

Monday, March 14, 2011

$\displaystyle{\left({6},{2}\right)},-{3}{x}+{4}{y}=-{5}$ Find the distance between the point and the line.

Given (6, 2), $\displaystyle-{3}{x}+{4}{y}=-{5}$

$\displaystyle-{3}{x}+{4}{y}=-{5}$

$\displaystyle-{3}{x}+{4}{y}+{5}={0}$

Let A=-3, B=4, C=5, x1=6, and y1=2

Find the distance between the point and the line using the formula

$\displaystyle{d}=\frac{{\left|{A}{x}_{{1}}+{B}{y}_{{1}}+{C}\right|}}{\sqrt{{{{A}}^{{2}}+{{B}}^{{2}}}}}$

$\displaystyle{d}=\frac{{\left|{\left(-{3}\right)}{\left({6}\right)}+{\left({4}\right)}{\left({2}\right)}+{5}\right|}}{{\sqrt{{{{\left(-{3}\right)}}^{{2}}+{\left({{4}}^{{2}}\right)}}}}}$

$\displaystyle{d}=\frac{{\left|-{18}+{8}+{5}\right|}}{\sqrt{{{9}+{16}}}}=\frac{{5}}{{5}}={1}$

The distance between the point and the line is 1 unit.

The Notes

Monday, March 14, 2011

$\displaystyle{\left({6},{2}\right)},-{3}{x}+{4}{y}=-{5}$ Find the distance between the point and the line.

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