Alex wants to fence in an area for a dog park. he has plotted three sides of the fenced area at the points e (0, 5), f (5, 5), and g (1, 1). he has 20 units of fencing. where could alex place point h so that he does not have to buy more fencing? (−3, 1) (−3, −1) (−5, 1) (−5, −1)

The answer is (-3,1)

Step-by-step explanation:

The points E (0,5) F (5,5) and G (1,1) lie on the graph such that the fence goes from E to F to G as shown in the graph.

From the graph we see that we only need to find the minimum distance between E and H, and between H and G.

The distances between E and the four points are:

Distance between E (0,5) and H (-3,1) is 5.

Distance between E (0,5) and H (-3,-1) is 6.7.

Distance between E (0,5) and H (-5,1) is 6.4.

Distance between E (0,5) and H (-5,-1) is 7.8.

The distances between G and the four points are:

Distance between E (1,1) and H (-3,1) is 4.

Distance between E (1,1) and H (-3,-1) is 4.47.

Distance between E (1,1) and H (-5,1) is 6.

Distance between E (1,1) and H (-5,-1) is 6.32.

From all the combinations we see that the Pint H (-3, 1) yields the shortest distance.  If Alex chooses point H (-3,1), he will have to use 19.5 units of fencing, thus he will not need to buy more.

-3,1

Step-by-step explanation:

graph them on a graphing calculator, press 2nd trace. select intersect. hit enter until it says answer. ti 84 works best

step-by-step explanation: