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Area of Triangle with Coordinates of Vertices Given

If x1 y1 ...

Question

If $∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣$ = $∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 b_{1} & b_{2} & b_{3} a_{1} & a_{2} & a_{3} \end{matrix} ∣ ∣ ∣ ∣$ then the two triangles whose vertices are $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ and $(a_{1}, b_{1}), (a_{2}, b_{2}), (a_{3}, b_{3})$ are

A

congruent

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B

similar

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C

equal in areas

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D

right angled triangles

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Solution

The correct option is C equal in areas
Area of triangle whose vertices are $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ is
$= ∣ ∣ ∣ ∣ ∣ \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$
And are of triangle whose vertices are $(a_{1}, b_{1}), (a_{2}, b_{2}), (a_{3}, b_{3})$ is
$= \frac{1}{2} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & 1 a_{2} & b_{2} & 1 a_{3} & b_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$
As $∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 b_{1} & b_{2} & b_{3} a_{1} & a_{2} & a_{3} \end{matrix} ∣ ∣ ∣ ∣$
$\Rightarrow ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 b_{1} & b_{2} & b_{3} a_{1} & a_{2} & a_{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$
$\Rightarrow ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & 1 a_{2} & b_{2} & 1 a_{3} & b_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$
$\Rightarrow$ both triangle have equal area

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Similar questions

∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 b_{1} & b_{2} & b_{3} a_{1} & a_{2} & a_{3} \end{matrix} ∣ ∣ ∣ ∣

then the two triangles whose vertices are

(x_{1}, y_{1}), (x_{2}, y_{2}), ((x_{3}, y_{3})

(a_{1}, b_{1}), (a_{2}, b_{2}), (a_{13}, b_{3}),

∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & 1 a_{2} & b_{2} & 1 a_{3} & b_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣

then the two triangles with vertices

(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

(a_{1}, b_{1}), (a_{2}, b_{2}), (a_{3}, b_{3})

∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & 1 a_{2} & b_{2} & 1 a_{3} & b_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣

, then the two triangles with vertices

(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

(a_{1}, b_{1})

(a_{2}, b_{2})

(a_{3}, b_{3})

must be congruent.

∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & 1 a_{2} & b_{2} & 1 a_{3} & b_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣

then two triangles with vertices

(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

(a_{1}, b_{1}), (a_{2}, b_{2}), (a_{3}, b_{3})

∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & 1 a_{2} & b_{2} & 1 a_{3} & b_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣

then the two triangles with vertices

(x_{1}, y_{1}), (x_{2}, y_{2}) (x_{3}, y_{3}) a n d (a_{1}, b_{1}), (a_{2}, b_{2}), (a_{3}, b_{3})

must be congruent . Is this true ?

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Area of Triangle with Coordinates of Vertices Given

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