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

Wednesday, November 19, 2014

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

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

$\displaystyle{3}{x}+{y}={1}$

$\displaystyle{3}{x}+{y}-{1}={0}$

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

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

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

$\displaystyle{d}=\frac{{\left|{\left({3}\right)}{\left({2}\right)}+{\left({1}\right)}{\left({3}\right)}+{\left(-{1}\right)}\right|}}{\sqrt{{{{3}}^{{2}}+{{1}}^{{2}}}}}=\frac{{\left|{6}+{3}-{1}\right|}}{\sqrt{{{10}}}}=\frac{{8}}{\sqrt{{10}}}={2.5}$

The distance between the point and the line is 2.5 units.

The Notes

Wednesday, November 19, 2014

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

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