43.6k views
0 votes
OVER 50 PTSSSSS!!!!!!!

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?

(0, 1)
(0, −2)
(1, 1)
(1, −2)

User Enoobong
by
7.6k points

2 Answers

2 votes

Answer: C, (1,1)

Step-by-step explanation: If you plot the three points given in the equation, you can then see that (1,1) is closest to the remaining points. Because it is the closest, the perimeter is shorter and Alex would not have to buy more fencing.: :)

User David Thomas
by
7.1k points
3 votes

Answer: That would be C

Step-by-step explanation: Let H's coordinates be (x, y).

We obtain using the distance formula,

EF is equal to ( 3 - 1)2 + (5 - 5)2 = ( 4 + 0) = 2 units.

FG = (9 + 16) = 5 units (i.e., 6 - 3)2 + (1 - 5)2).

→ GH = √{(x - 6)² + (y - 1)²}

→ HE = √{(x - 1)² + (y - 5)²}

since,

16 units = EF + FG + GH + HE

→ 2 + 5 + √{(x - 6)² + (y - 1)²} + √{(x - 1)² + (y - 5)²} = 16

9 units are equal to (x - 6)2 + (y - 1)2 + (x - 1)2 + (y - 5)2.

utilizing the options available to us now,

A) (0, 1) (0, 1)

→ √(36) + √(1 + 25) = 6 + √26 ≠ 9 units.

B) (0, - 2) (0, - 2)

→ √(36 + 9) + √(1 + 49) = √45 + √50 ≠ 9 units .

C) (1 , 1) (1 , 1)

→ √(25 + 0) + √(0 + 16) = 5 + 4 = 9 = 9 units .

Alex should then position Point H at (1, 1).

User David Folkner
by
7.6k points