The Notes: `(4, -4), y = -2x - 3` Find the distance between the point and the line.

Friday, March 1, 2013

The formula to determine the distance from a point (xo,yo) 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.

$\displaystyle{y}=-{2}{x}-{3}$

$\displaystyle{2}{x}+{y}+{3}={0}$

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

$\displaystyle{d}=\frac{{\left|{2}{x}_{{o}}+{1}{y}_{{o}}+{3}\right|}}{\sqrt{{{{2}}^{{2}}+{{1}}^{{2}}}}}$

$\displaystyle{d}=\frac{{\left|{2}{x}_{{o}}+{y}_{{o}}+{3}\right|}}{\sqrt{{5}}}$

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

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

$\displaystyle{d}=\frac{{7}}{\sqrt{{5}}}$

$\displaystyle{d}=\frac{{7}}{\sqrt{{5}}}\cdot\frac{\sqrt{{5}}}{\sqrt{{5}}}$

$\displaystyle{d}=\frac{{{7}\sqrt{{5}}}}{{5}}$

Therefore, the point (4,-4) is $\displaystyle\frac{{{7}\sqrt{{5}}}}{{5}}$ units away from the line y=-2x-3.

The Notes