EXAMPLE 1 Standardized Test Practice

SOLUTION

Let ( x1, y1 ) = ( –3, 5) and ( x2, y2 ) = ( 4, – 1 ).

= (4 – (–3))2 + (– 1 – 5)2

= 49 + 36 = 85

( x2 – x1 )2 + ( y2 – y1)2=d

ANSWER

The correct answer is C.

EXAMPLE 2 Classify a triangle using the distance formula

ANSWER

Because BC = AC, ∆ABC is isosceles.

Classify ∆ABC as scalene, isosceles, or equilateral.

AB = (7 – 4)2 + (3 – 6)2 = 18 = 3 2

BC = (2 – 7)2 + (1 – 3)2 = 29

AC = (2 – 4)2 + (1 – 6)2 = 29

GUIDED PRACTICE for Examples 1 and 2

SOLUTION

Let ( x1, y1 ) = ( 3, – 3) and ( x2, y2 ) = (–1, 5).

1. What is the distance between (3, – 3) and (– 1, 5)?

= (–1 – 3)2 + (5 – (–3))2

= 16 + 64

( x2 – x1 )2 + ( y2 – y1)2=d

= 80

4 5

ANSWER The distance between (3, –3) and (–1, 5) is

GUIDED PRACTICE

2. The vertices of a triangle are R(– 1, 3), S(5, 2), and T(3, 6). Classify ∆RST as scalene, isosceles, or equilateral.

R – 1,3

S 5, 2

T 3, 6

for Examples 1 and 2

GUIDED PRACTICE

SOLUTION

for Examples 1 and 2

ST = (3 – 5)2 + (6 – 2)2 = 20

TR = (–1 –(–3)2 + (3 – 6)2 = 25

= 5

RS = (5 – (–1)2 + (2 – 3)2 = 36 = 6

ANSWER Because RS ≠ ST ≠ TR, so RSTis an scalene triangle.