Answer:
The values of y such that the barn is 70 yards from the well are y=76 or y=-64
Step-by-step explanation:
Distance in the Plane
Given two points in a rectangular system of coordinates (a,b) and (c,d) the distance measured between them is calculated with the formula
[tex]d=\sqrt{(c-a)^2+(d-b)^2}[/tex]
The barn is located at (4,6) and the well is located at (4,y). The value of y must be calculated in such a way the distance between the barn and the well is 70 yards. Thus, the equation to solve is
[tex]70=\sqrt{(4-4)^2+(y-6)^2}[/tex]
Operating
[tex]70=\sqrt{(y-6)^2}[/tex]
When taking the square root we must be careful for it has two signs:
[tex]70=\pm (y-6)[/tex]
Rearranging
[tex]y-6=\pm 70[/tex]
which yields to these solutions
[tex]y-6=70[/tex]
[tex]\boxed{y=76}[/tex]
And also
[tex]y-6=-70[/tex]
[tex]\boxed{y=-64}[/tex]
The values of y such that the barn is 70 yards from the well are y=76 yards or y=-64 yards
