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The Distance Formula

You know that thedistance $AB$between two points in a plane withCartesiancoordinates$A\left(x_1,y_1\right)$and$B\left(x_2,y_2\right)$由以下公式给出:

$AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The distance formula is really just thePythagorean Theorem在伪装。

To calculate the distance$AB$between point$A\left(x_1,y_1\right)$and$B\left(x_2,y_2\right)$, first draw a right triangle which has the segment$\overline{AB}$as its hypotenuse.

If the lengths of the sides are$a$and$b$, then by the Pythagorean Theorem,

${\left(AB\right)}^2={\left(AC\right)}^2+{\left(BC\right)}^2$

Solving for the distance$AB$, we have:

$AB=\sqrt{{\left(AC\right)}^2+{\left(BC\right)}^2}$

Since$AC$is a horizontal distance, it is just the difference between the$x$-coordinates:$\left|\left(x_2-x_1\right)\right|$. Similarly,$BC$is the vertical distance$\left|\left(y_2-y_1\right)\right|$.

Since we're squaring these distances anyway (and squares are always non-negative), we don't need to worry about those absolute value signs.

$AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

In the above example, we have:

$A\left(x_1,y_1\right)=\left(-1,0\right),B\left(x_2,y_2\right)=\left(2,7\right)$

so

$\begin{array}{l}AB=\sqrt{{\left(2-\left(-1\right)\right)}^2+{\left(7-0\right)}^2}\\ =\sqrt{3^2+7^2}\\ =\sqrt{9+49}\\ =\sqrt{58}\end{array}$

or approximately$7.6$units.