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

Thursday, December 17, 2009

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

The formula to determine the distance from a point $\displaystyle{\left({x}_{{o}},{y}_{{o}}\right)}$ to a line Ax+By+C=0 is:

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

To apply, set one side of the given equation equal to zero.

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

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

Then, plug-in the coefficients of x and y, as well as the constant to the formula.

$\displaystyle{d}=\frac{{\left|{\left(-{2}\right)}{x}_{{o}}+{1}{y}_{{o}}+{\left(-{2}\right)}\right|}}{\sqrt{{{{\left(-{2}\right)}}^{{2}}+{{1}}^{{2}}}}}$

$\displaystyle{d}=\frac{{\left|-{2}{x}_{{o}}+{y}_{{o}}-{2}\right|}}{\sqrt{{5}}}$

And, plug-in the given point (2,1).

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

$\displaystyle{d}=\frac{{\left|-{5}\right|}}{\sqrt{{5}}}$

$\displaystyle{d}=\frac{{5}}{\sqrt{{5}}}$

$\displaystyle{d}=\frac{{5}}{\sqrt{{5}}}\cdot\frac{\sqrt{{5}}}{\sqrt{{5}}}$

$\displaystyle{d}=\frac{{{5}\sqrt{{5}}}}{{5}}$

$\displaystyle{d}=\sqrt{{5}}$

Therefore, the distance between the point (2,1) and the line is $\displaystyle\sqrt{{5}}$ units.

The Notes

Thursday, December 17, 2009

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

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