The Notes: `(1, -3), y = 2x -5` Find the distance between the point and the line.

Saturday, March 8, 2014

$\displaystyle{\left({1},-{3}\right)},{y}={2}{x}-{5}$ Find the distance between the point and the line.

Given (1, -3) $\displaystyle{y}={2}{x}-{5}$

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

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

Let A=-2, B=1, C=5 and x1=1, y1=-3

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

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

$\displaystyle{d}=\frac{{\left|{\left(-{2}\right)}{\left({1}\right)}+{\left({1}\right)}{\left(-{3}\right)}+{5}\right|}}{\sqrt{{{{\left(-{2}\right)}}^{{2}}+{{\left({1}\right)}}^{{2}}}}}$

$\displaystyle{d}=\frac{{\left|-{2}-{3}+{5}\right|}}{\sqrt{{{4}+{1}}}}=\frac{{0}}{\sqrt{{5}}}={0}$

The distance between the point and the line is 0.

The Notes

Saturday, March 8, 2014

$\displaystyle{\left({1},-{3}\right)},{y}={2}{x}-{5}$ Find the distance between the point and the line.

No comments:

Post a Comment

What are hearing tests?