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.

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 :

.

Comparing it with given D :

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

Distance between Elkin exit to the Mt. Airy exit.

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

Hence ,this is the required solution .

dilation scale factor 2: starting with δabc, draw the dilation image of the triangle with a center at the origin and a scale factor of two. notice that every coordinate of the original triangle has been multiplied by the scale factor (x2). dilation with scale factor 2, multiply by 2.

the answer is  

a.

4,624π mm2

step-by-step explanation: