CK-12 Geometry - Second Edition

3.6. The Distance Formula http://www.ck12.org

d=

√

(− 10 − 4 )^2 +( 3 + 2 )^2

=

√

(− 14 )^2 +( 52 ) Distances are always positive!

=

√

196 + 25

=

√

221 ≈ 14. 87 units

Example 2:The distance between two points is 4 units. One point is (1, -6). What is the second point? You may
assume that the second point is made up of integers.

Solution:We will still use the distance formula for this problem, however, we knowdand need to solve for(x 2 ,y 2 ).

4 =

√

( 1 −x 2 )^2 +(− 6 −y 2 )^2

16 = ( 1 −x 2 )^2 +(− 6 −y 2 )^2

At this point, we need to figure out two square numbers that add up to 16. The only two square numbers that add up
to 16 are 16+0.

16 = ( 1 −x 2 )^2

︸ ︷︷ ︸

42

+(− 6 −y 2 )^2

︸ ︷︷ ︸

02

or 16 = ( 1 −x 2 )^2

︸ ︷︷ ︸

02

+(− 6 −y 2 )^2

︸ ︷︷ ︸

42

1 −x 2 =± 4 − 6 −y 2 = 0 1 −x 2 = 0 − 6 −y 2 =± 4

−x 2 =−5 or 3 and −y 2 = 6 or −x 2 =− 1 and y 2 =10 or 2

x 2 =5 or− 3 y 2 =− 6 x 2 = 1 y 2 =−10 or− 2

Therefore, the second point could have 4 possibilities: (5, -6), (-3, -6), (1, -10), and (1, -2).

Shortest Distance between a Point and a Line

We know that the shortest distance between two points is a straight line. This distance can be calculated by using
the distance formula. Let’s extend this concept to the shortest distance between a point and a line.

Just by looking at a few line segments fromAto linel, we can tell that the shortest distance between a point and a
line is theperpendicular linebetween them. Therefore,ADis the shortest distance betweenAand linel.

Putting this onto a graph can be a little tougher.

Example 3:Determine the shortest distance between the point (1, 5) and the liney=^13 x−2.