Intermediate Algebra (11th edition)

By the Pythagorean theorem, the square of the length of the hypotenuse dof the

right triangle in FIGURE 8is equal to the sum of the squares of the lengths of the two

legs aand b.

The length ais the difference between the y-coordinates of the endpoints. Since

the x-coordinate of both points in FIGURE 8is the side is vertical, and we can find

aby finding the difference between the y-coordinates. We subtract from 3 to get

a positive value for a.

Similarly, we find bby subtracting from 3.

Now substitute these values into the equation.

Let and.

Apply the exponents.

Add.

Choose the principal root.

We choose the principal root, since distance cannot be negative. Therefore, the dis-

tance between and is

NOTE It is customary to leave the distance in simplified radical form. Do not use a

calculator to get an approximation, unless you are specifically directed to do so.

This result can be generalized. FIGURE 9 shows the two points and

The distance abetween and is given by

and the distance bbetween and is given by

From the Pythagorean theorem, we obtain the following.

Choosing the principal square root gives the distance formula.

d^2 = 1 x 2 - x 122 + 1 y 2 - y 122

d^2 = a^2 +b^2

b= |y 2 - y 1 |.

1 x 2 , y 22 1 x 2 , y 12

a= |x 2 - x 1 |,

1 x 2 , y 22. 1 x 1 , y 12 1 x 2 , y 12

1 x 1 , y 12

1 - 5, 3 2 1 3, - 42 2113.

d= 2113

d^2 = 113

d^2 = 49 + 64

d 2 = 72 + 82 a= 7 b= 8

d^2 = a^2 +b^2

b= 3 - 1 - 52 = 8

- 5

a= 3 - 1 - 42 = 7

- 4

- 5,

a^2 b^2 d^2

SECTION 8.3 Simplifying Radical Expressions 449

d b

(x 1 , y 1 ) a (x 2 , y 1 )

(x 2 , y 2 )

y

0

x

FIGURE 9

Distance Formula

The distance dbetween the points and is

d 21 x 2 x 122  1 y 2 y 122.

1 x 1 , y 12 1 x 2 , y 22

x

y

0

(–5, 3)

(–5, – 4)

(3, – 4)

d

a

b

FIGURE 8