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

Friday, June 27, 2014

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

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

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

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

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

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

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

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

$\displaystyle{d}=\frac{{\left|-{6}+{2}+{1}\right|}}{\sqrt{{{4}+{1}}}}=\frac{{3}}{\sqrt{{5}}}={1.3}$

The distance between the point and the line is 1.3 units.

The Notes

Friday, June 27, 2014

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

No comments:

Post a Comment

What are hearing tests?