CK-12 Geometry Concepts

3.10. Distance Formula in the Coordinate Plane 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 B

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.

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).

Example C

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

First, graph the line and point. Second determine the equation of the perpendicular line. The opposite sign and
reciprocal of^13 is -3, so that is the slope. We know the line must go through the given point, (1, 5), so use that to find
they−intercept.

y=− 3 x+b

5 =− 3 ( 1 )+b The equation of the line isy=− 3 x+ 8.

8 =b

Next, we need to find the point of intersection of these two lines. By graphing them on the same axes, we can see
that the point of intersection is (3, -1), the green point.

Finally, plug (1, 5) and (3,-1) into the distance formula to find the shortest distance.