The coordinates of E in Quadrant lll is (3 - [tex]2\sqrt{3}[/tex], -4)
Given,
m < F = 90°
E (x, y) in quadrant lll
F(-3, 4), G(-3, 2)
FG = The longer leg.
FG = [tex]\sqrt{(-3-(-3)^{2} )+(2-(-4)^{2} }[/tex] = [tex]\sqrt{36}[/tex] = 6
FE = [tex]\frac{FG}{\sqrt{3} }[/tex] = [tex]\frac{6}{\sqrt{3} }[/tex] = 2[tex]\sqrt{3}[/tex]
GE = 2FE = 2 × 2√3 = 4√3
ΔEFG is a 30° - 60° - 90° triangle with the longest leg FG = 6.
We apply the 30° - 60° - 90° triangle theorem to determine the short side FE and the hypotenuse GE.
l = 4√3
h = 2√3
[tex]\sqrt{(x-(-3))^{2} +(y-(-4))^{2} }[/tex] = (2√3)²
[tex]\sqrt{(x-(-3))^{2} +(y-2)^{2} }[/tex] = (4√3)²
∴ (x + 3)² + ( y + 4)² = 12
(x + 3)² + (y - 2)² = 48
Solving this, we get:
Coordinates of E as (3 - 2√3, -4)
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