Question

The vertices of triangle XYZ are X(–1, 5), Y(–1, –2), and Z(–5, –2). What is the approximate distance between vertex X and vertex Z?

Answer 1

Explanation:

From distance formula

(x

2

−x

1

)+(y

2

−y

1

)+(z

2

−z

1

)

AB=

1+4+1

=

6

AC=

36+5+9

=

56

=3

6

AC

AB

=

EC

BE

⇒

3

6

6

=

EC

BE

⇒

EC

BE

=

3

1

So now we can find point E

E=(

m

1

+m

2

x

1

x

2

+m

2

m

1

),(

m

1

+m

2

m

1

y

2

+m

2

y

1

),(

m

1

m

2

m

1

z

2

+m

2

z

1

)

E=(

4

−5+6

,

4

2+3

,

4

−6−6

)=E(

4

1

,

4

5

,−3)

Now we can find distance AE

AE=

(1−

4

1

)

2

+(1−

4

5

)

2

+(−3+3)

2

AE=

16

9

+8

16

1

+0

AE=

4

3

10