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{{ printedBook.courseTrack.name }} {{ printedBook.name }} If $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$ are two points in a coordinate plane, the distance between them, $AB,$ can be found.

The horizontal and vertical distance between the points can be thought of as $Δx$ and $Δy,$ respectively.

As can be seen in the diagram above, $AB,$ which gives the distance between points $A$ and $B,$ is the hypotenuse of the right triangle whose legs are $Δx$ and $Δy.$ An equation can be created for $AB$ using the Pythagorean Theorem, $a_{2}+b_{2}=c_{2}.$ $(Δx)_{2}+(Δy)_{2}=(AB)_{2}$ $Δx$ and $Δy$ are the differences between the coordinates of the points. They can be written as $Δx=∣x_{2}−x_{1}∣$ and $Δy=∣y_{2}−y_{1}∣.$ Here, the absolute values are included to ensure the distances are positive. Rearranging the equation and substituting these values gives the following.

$(AB)_{2}=(Δx)_{2}+(Δy)_{2}$

$(AB)_{2}=(∣x_{2}−x_{1}∣)_{2}+(∣y_{2}−y_{1}∣)_{2}$

Because squares are always positive, the absolute value bars can be removed. Then, $AB$ can be isolated.

$(AB)_{2}=(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2}$

SqrtEqn$LHS =RHS $

$AB=±(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2} $

Q.E.D.