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)

..

step-by-step explanation:

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.