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)

(-3,1)

Step-by-step explanation:

First we need to know the distance between E and F, and the distance betweem F and G:

d(E,F) = sqrt((5-0)^2 + (5-5)^2) = 5

d(F,G) = sqrt((5-1)^2 + (5-1)^2) = 5.6569

The other distances are d(G,H) and d(H,E).

So, if we choose the point H = (-3,1), we have:

d(G,H) = sqrt((1+3)^2 + (1-1)^2) = 4

d(H,E) = sqrt((-3-0)^2 + (1-5)^2) = 5

Total fencing = 5 + 5.6569 + 4 + 5 = 19.6569 < 20

This point can be chosen.

If we take H = (-3,-1):

d(G,H) = sqrt((1+3)^2 + (1+1)^2) = 4.4721

d(H,E) = sqrt((-3-0)^2 + (-1-5)^2) = 6.7082

Total fencing = 5 + 5.6569 + 4.4721 + 6.7082 = 21.8372 > 20

This point can't be chosen.

If we take H = (-5,1):

d(G,H) = sqrt((1+5)^2 + (1-1)^2) = 6

d(H,E) = sqrt((-5-0)^2 + (1-5)^2) =6.4031

Total fencing = 5 + 5.6569 + 6 + 6.4031 = 23.06 > 20

This point can't be chosen.

If we take H = (-5,-1):

d(G,H) = sqrt((1+5)^2 + (1+1)^2) = 6.3246

d(H,E) = sqrt((-5-0)^2 + (-1-5)^2) = 7.8102

Total fencing = 5 + 5.6569 + 6.3246 + 7.8102 = 24.7917 > 20

This point can't be chosen.

So the point Alex should place point H is (-3,1)

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3