The Notes: `(1, 4), y = 4x + 2` Find the distance between the point and the line.

Sunday, April 5, 2009

Take note that 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.

$\displaystyle{y}={4}{x}+{2}$

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

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

$\displaystyle{d}=\frac{{\left|-{4}{x}_{{o}}+{1}{y}_{{o}}+{\left(-{2}\right)}\right|}}{\sqrt{{{{\left(-{4}\right)}}^{{2}}+{{1}}^{{2}}}}}$

$\displaystyle{d}=\frac{{\left|-{4}{x}_{{o}}+{y}_{{o}}-{2}\right|}}{\sqrt{{17}}}$

And, plug-in the given point (1,4).

$\displaystyle{d}=\frac{{\left|-{4}{\left({1}\right)}+{4}-{2}\right|}}{\sqrt{{17}}}$

$\displaystyle{d}=\frac{{\left|-{2}\right|}}{\sqrt{{17}}}$

$\displaystyle{d}=\frac{{2}}{\sqrt{{17}}}$

$\displaystyle{d}=\frac{{2}}{\sqrt{{17}}}\cdot\frac{\sqrt{{17}}}{\sqrt{{17}}}$

$\displaystyle{d}=\frac{{{2}\sqrt{{17}}}}{{17}}$

Therefore, the distance between the point (1,4) and the line y=4x+2 is $\displaystyle\frac{{{2}\sqrt{{17}}}}{{17}}$ units.

The Notes