Question

1. Three lines x + 3y = 6. 2x - 3y = 12 and x = 0 represent the sides of a triangular beautiful garden. Find the intersection of these three lines geometrically and if all measurements are km, find the area of ​​the garden. What kind of treatment should be expected from the public who come to such a beautiful garden?

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Answer 1

Answer:

18 km²

Step-by-step explanation:

$d = \sqrt{(x_{2} - x_{1})^2 +(y_{2} -y_{1} )^2}$

P = a + b + c

s = $\frac{P}{2}$

$A_{ABC}$ = $\sqrt{s(s-a)(s-b)(s-c)}$

~~~~~~~~~~

2x - 3y = 12 ⇒ y = $\frac{2}{3}$ x - 4

x + 3y = 6 ⇒ y = $-\frac{1}{3}$ x + 2

A(0,- 4), B(0, 2), C(6, 0)

c = AB = √[(0)² + (- 4 - 2)²] = 6 km

a = BC = √(6² + 2²) = √40 = 2√10 ≈ 6.33 km

b = AC = √[6² + (- 4)²] = √52 =2√13 ≈ 7.21 km

P = 19.54 km

s = 9.77 km

A = $\sqrt{9.77(9.77-6.33)(9.77-7.21)(9.77-6)}$ ≈ 18 km²