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Area of a Triangle Given Its Vertices

a cos ϕ 1, b ...

Question

$(a cos ϕ_{1}, b sin ϕ_{1}), (a cos ϕ_{2}, b sin ϕ_{2})$ and $(a cos ϕ_{3}, b sin ϕ_{3})$

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Solution

Area of the triangle having coordinates $(a cos ϕ_{1}, b sin ϕ_{1}), (a cos ϕ_{2}, b sin ϕ_{2}), (a cos ϕ_{3}, b sin ϕ_{3})$ is given by
$\frac{1}{2} ∣ ∣ ∣ \begin{matrix} a cos ϕ_{1} & a cos ϕ_{2} & a cos ϕ_{3} & a cos ϕ_{1} b sin ϕ_{1} & b sin ϕ_{2} & b sin ϕ_{3} & b sin ϕ_{1} \end{matrix} ∣ ∣ ∣$
$= \frac{1}{2} \times (a cos ϕ_{1} b sin ϕ_{2} + a cos ϕ_{2} b sin ϕ_{3} + a cos ϕ_{3} b sin ϕ_{1} - a cos ϕ_{2} b sin ϕ_{1} - a cos ϕ_{3} b sin ϕ_{2} - a cos ϕ_{1} b sin ϕ_{3})$
$= \frac{a b}{2} \times (sin (ϕ_{2} - ϕ_{1}) + sin (ϕ_{3} - ϕ_{2}) + sin (ϕ_{1} - ϕ_{3}))$

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