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 :

$(\dfrac{h-5}{2},\dfrac{k-2}{2})$ .

Comparing it with given D :

$\dfrac{h-5}{2}=-1\\\\h=3$ $\dfrac{k-2}{2}=4\\\\k=10$

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

Distance between Elkin exit to the Mt. Airy exit.

$D=\sqrt{(5-3)^2+(-2-10)^2}\\\\D=12.17\ units$

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

Hence ,this is the required solution .