CK-12 Geometry-Concepts

http://www.ck12.org Chapter 3. Parallel and Perpendicular Lines

Example A

Find the distance between (4, -2) and (-10, 3).

Plug in (4, -2) for(x 1 ,y 1 )and (-10, 3) for(x 2 ,y 2 )and simplify.

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.