The Notes: `(-1, 2), 5x + 3y = -4` Find the distance between the point and the line.

Tuesday, February 3, 2015

$\displaystyle{\left(-{1},{2}\right)},{5}{x}+{3}{y}=-{4}$ Find the distance between the point and the line.

Given (-1, 2) $\displaystyle{5}{x}+{3}{y}=-{4}$

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

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

Let A=5, B=3, C=4, x1=-1, and y1=2

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

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

$\displaystyle{d}=\frac{{\left|{\left({5}\right)}{\left(-{1}\right)}+{\left({3}\right)}{\left({2}\right)}+{4}\right|}}{\sqrt{{{{5}}^{{2}}+{{3}}^{{2}}}}}$

$\displaystyle{d}=\frac{{\left|-{5}+{6}+{4}\right|}}{\sqrt{{{34}}}}=\frac{{5}}{\sqrt{{{34}}}}={.8575}$

The distance between the point and the line is .8575 units.

The Notes

Tuesday, February 3, 2015

$\displaystyle{\left(-{1},{2}\right)},{5}{x}+{3}{y}=-{4}$ Find the distance between the point and the line.

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