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

Thursday, October 4, 2012

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

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

$\displaystyle{4}{x}-{5}{y}=-{2}$

$\displaystyle{4}{x}-{5}{y}+{2}={0}$

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

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

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

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

$\displaystyle{d}=\frac{{\left|{8}+{15}+{2}\right|}}{\sqrt{{{16}+{25}}}}=\frac{{25}}{\sqrt{{{41}}}}={3.9}$

The distance between the point and the line is 3.9 units.

The Notes

Thursday, October 4, 2012

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

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