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 :
[tex](\dfrac{h-5}{2},\dfrac{k-2}{2})[/tex] .
Comparing it with given D :
[tex]\dfrac{h-5}{2}=-1\\\\h=3[/tex] [tex]\dfrac{k-2}{2}=4\\\\k=10[/tex]
Therefore , the location of Mt. Airy exit is ( 3, 10 ) .
Distance between Elkin exit to the Mt. Airy exit.
[tex]D=\sqrt{(5-3)^2+(-2-10)^2}\\\\D=12.17\ units[/tex]
Therefore , distance between Elkin exit to the Mt. Airy exit is 12.17 units .
Hence ,this is the required solution .
