The Notes: `(-2, 6), y = -x + 5` Find the distance between the point and the line.

Wednesday, January 1, 2014

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.

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

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

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

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

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

And, plug-in the given point (-2,6).

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

$\displaystyle{d}=\frac{{\left|-{1}\right|}}{\sqrt{{2}}}$

$\displaystyle{d}=\frac{{1}}{\sqrt{{2}}}$

$\displaystyle{d}=\frac{{1}}{\sqrt{{2}}}\cdot\frac{\sqrt{{2}}}{\sqrt{{2}}}$

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

Therefore, the distance between the point (-2,6) and the line y=-x+5 is $\displaystyle\frac{\sqrt{{2}}}{{2}}$ units.

The Notes