On a coordinate plane, triangle A B C has points (negative 2, negative 1), (negative 2, negative 3), (negative 5, negative 1). T
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The usual substitutions can be used:
x = r·cos(θ)
y = r·sin(θ)
These give
2r·cos(θ) -2r·sin(θ) = 3
Solving for r, we get the desired equation:
r = 3/(2·(cos(θ) - sin(θ)))
x = r·cos(θ)
y = r·sin(θ)
These give
2r·cos(θ) -2r·sin(θ) = 3
Solving for r, we get the desired equation:
r = 3/(2·(cos(θ) - sin(θ)))