Alex wants to fence in an area for a dog park. he has plotted three sides of the fenced area at the points e (1, 5), f (3, 5), and g (6, 1). he has 16 units of fencing. where could alex place point h so that he does not have to buy more fencing? a.(0,1) b. (0,-2) c.(1,1) d.(1,-2)

C (1,1)

Step-by-step explanation:

Let's figure out how much fencing is has already used with the given 3 points.

Graph the points to help visualize.

The distance between (1,5) and (3,5) is a horizontal one so that is easy to find, that is 2.  

The distance between (3,5) and (6,1) is a bit harder (because it isn't a vertical or just horizontal distance).  So let's use the distance formula: sqrt((6-3)^2+(5-1)^2)=sqrt(3^2+4^2)=sqrt(9+16)=sqrt(25)=5.  

So we have used 2+5 of our fencing. We have 16-7=9 units of fencing left to use.

I think I'm going to plot these points in the choices on my graph and see which is the least furthest away from the other points. The answer would make sense to be (1,1).  Bet let's just test it.

Assume the answer is C.

The distance from (6,1) to (1,1) is 5.

The distance from (1,1) to (1,5) is 4.

That is 9 which is what we had left over.

So actually this question was way easier than I made it but I will leave it here because it is still good work and it shows that choice C actually does lie within the conditions of the question.

The closest point in the choices to all the other points in the problem was (1,1) so that would make it the answer (assuming the question maker didn't make a mistake :p)

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step-by-step explanation: