What is the Distance Formula?

The distance between two points on a coordinate plane can found by using the distance formula.

To understand and discover the distance formula you must use the Pythagorean Theorem.  If you have a right triangle, then the square on the hypotenuse is equal to the sum of the squares of the other two sides. 

a^{2}+b^{2}=c^{2}

The converse is also true.  If a^{2}+b^{2}=c^{2} , then the triangle is a right triangle.

Find the distance between two points \left (x_{1},y_{1} \right ) and \left (x_{2},y_{2} \right ).

The Length of the Vertical side is \left | y_{2}-y_{1} \right | or \left | y_{1}-y_{2} \right |, so b = \left | y_{2}-y_{1} \right |.

The Length of the Horizontal side is \left | x_{2}-x_{1} \right |. so a = \left | x_{2}-x_{1} \right |.

The hypotenuse of the right triangle is the distance between the two points,  so c = d.

c^{2}=a^{2}+b^{2}

d^{2}=\left | x_{2}-x_{1} \right |^{2}+\left | y_{2}-y_{1} \right |^{2} d=\sqrt{\left ( x_{2} -x_{1}\right )^{2}+\left ( y_{2} -y_{1}\right )^{2}}

Examples:

Find the distance between the following two points:

1)  \left ( 6,-3 \right ),\left ( 6,5 \right )                d=\sqrt{\left ( 6-6 \right )^{2}+\left ( -3-5 \right )^{2}}

                                         d=\sqrt{0+\left ( -8 \right )^{2}} 

                                        d=\sqrt{64} 

                                        d=8 

2.  \left ( -3,-1 \right ),\left ( 2,-1 \right )           d=\sqrt{\left ( -3-2 \right )^{2}+\left ( -1-\left ( -1 \right ) \right )^{2}} 

                                         d={\sqrt{25+0}}

                                         d=5 

Determine if the triangle formed by joining these three points is a right triangle.

3.  \left ( 4,5 \right ),\left ( 0,2 \right ),\left ( 4,2 \right ) 

a)  \left ( 4,5 \right ),\left ( 0,2 \right )                              b)  \left ( 4,5 \right ),\left ( 4,2 \right )                           c)  \left ( 0,2\right ),\left ( 4,2 \right )

d=\sqrt{\left ( 4^{2} +3^{2}\right )}                               d=\sqrt{\left ( 0^{2} +3^{2}\right )}                           d=\sqrt{\left ( 4^{2} +0^{2}\right )}

d=\sqrt{16+9}                                    d=\sqrt{9}                                        d=\sqrt{16}

d^{2}=25                                           d^{2}=9                                           d^{2}=16 

25=16+9    Yes, these three points connect to make a right triangle.

4.  \left ( -1,1 \right ),\left ( 9,4 \right ),\left ( 9,1 \right )

a)  \left ( -1,1 \right ),\left ( 9,4 \right )                           b)  \left ( -1,1 \right ),\left ( 9,1 \right )                            c)  \left ( 9,4 \right ),\left ( 9,1 \right )

 d=\sqrt{\left (-1-9 \right )^{2}+\left ( 1-4 \right )^{2}}              d=\sqrt{\left (-1-9 \right )^{2}+\left ( 1-1 \right )^{2}}             d=\sqrt{\left (9-9 \right )^{2}+\left ( 4-1 \right )^{2}}

d=\sqrt{10^{2}+3^{2}}                                                         d=\sqrt{10^{2}+0^{2}}                                                        d=\sqrt{0^{2}+3^{2}} 

d^{2}=109                                                                        d^{2}=100                                                                        d^{2}=9 

109=100+9            Yes, these three points connect to make a right triangle.

Filed under: Pre-requisites for Pre-Calculus

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