The Notes: `(-1, -5), 6x + 3y = 3` Find the distance between the point and the line.

Sunday, July 20, 2014

$\displaystyle{\left(-{1},-{5}\right)},{6}{x}+{3}{y}={3}$ Find the distance between the point and the line.

Given (-1, -5) $\displaystyle{6}{x}+{3}{y}={3}$

$\displaystyle{6}{x}+{3}{y}={3}$

$\displaystyle{6}{x}+{3}{y}-{3}={0}$

$\displaystyle{2}{x}+{y}-{1}={0}$

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

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

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

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

$\displaystyle{d}=\frac{{\left|-{2}-{5}-{1}\right|}}{\sqrt{{{5}}}}=\frac{{8}}{\sqrt{{5}}}={3.6}$

The distance between the point and the line is 3.6 units.

The Notes

Sunday, July 20, 2014

$\displaystyle{\left(-{1},-{5}\right)},{6}{x}+{3}{y}={3}$ Find the distance between the point and the line.

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