The Notes: `(-2, 8), y = -3x + 2` Find the distance between the point and the line.

Friday, December 25, 2009

$\displaystyle{\left(-{2},{8}\right)},{y}=-{3}{x}+{2}$ Find the distance between the point and the line.

The distance between a point and line will be measured by finding the segment that is perpendicular to the line and goes through the point. The formula to find this distance is:

$\displaystyle\text{distance}{\left({a}{x}+{b}{y}+{c}={0},{\left({x}{1},{y}{1}\right)}\right)}=\frac{{\left|{a}{x}{1}+{b}{y}{1}+{c}\right|}}{\sqrt{{{{a}}^{{2}}+{{b}}^{{2}}}}}$

So, in our problem, we need to set the equation of our line equal to 0, so it ends up being $\displaystyle{3}{x}+{y}-{2}={0}$

We can now plug our numbers into the equation as follows:

$\displaystyle\text{distance}{\left({3}{x}+{y}-{2}={0},{\left(-{2},{8}\right)}\right)}=\frac{{\left|{3}{\left(-{2}\right)}+{1}{\left({8}\right)}-{2}\right|}}{\sqrt{{{{3}}^{{2}}+{{1}}^{{2}}}}}$

Which leaves us with an answer of $\displaystyle\frac{{0}}{\sqrt{{{10}}}}={0}$

The Notes

Friday, December 25, 2009

$\displaystyle{\left(-{2},{8}\right)},{y}=-{3}{x}+{2}$ Find the distance between the point and the line.

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