Question

On highway 77, the Dobson exit is halfway between the Mt. Airy exit and the Elkin exit. If the Elkin exit is located at (-5,-2) and the Dobson exit is located at (-1,4), how far is it from the Elkin exit to the Mt. Airy exit.

Answer 1

Given :

The Dobson exit is halfway between the Mt. Airy exit and the Elkin exit. If the Elkin exit is located at (-5,-2) and the Dobson exit is located at (-1,4).

To Find :

How far is it from the Elkin exit to the Mt. Airy exit.

Solution :

E(-5,-2) , D(-1 ,4 ) .

It is also given that Dobson exit is halfway between the Mt. Airy exit and the Elkin exit.

Let , location of Elkin exit is A(h,k) .

So , point D in terms of A and E is given by :

`((h-5)/(2),(k-2)/(2))` .

Comparing it with given D :

`(h-5)/(2)=-1\\\\h=3`
`(k-2)/(2)=4\\\\k=10`

Therefore , the location of Mt. Airy exit is ( 3, 10 ) .

Distance between Elkin exit to the Mt. Airy exit.

`D=√((5-3)^2+(-2-10)^2)\\\\D=12.17\ units`

Therefore , distance between Elkin exit to the Mt. Airy exit is 12.17 units .

Hence ,this is the required solution .