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

Tuesday, January 1, 2013

Take note that 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.

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

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

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

$\displaystyle{d}=\frac{{\left|{1}{x}_{{0}}+{\left(-{1}\right)}{y}_{{o}}+{2}\right|}}{\sqrt{{{{1}}^{{2}}+{{\left(-{1}\right)}}^{{2}}}}}$

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

And, plug-in the given point (2,1).

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

$\displaystyle{d}=\frac{{\left|{3}\right|}}{\sqrt{{2}}}$

$\displaystyle{d}=\frac{{3}}{\sqrt{{2}}}$

$\displaystyle{d}=\frac{{3}}{\sqrt{{2}}}\cdot\frac{\sqrt{{2}}}{\sqrt{{2}}}$

$\displaystyle{d}=\frac{{{3}\sqrt{{2}}}}{{2}}$

Therefore, the distance between the point (2,1) and the line y=x+2 is $\displaystyle\frac{{{3}\sqrt{{2}}}}{{2}}$ units.

The Notes