If A x 1- y 1- B x 2- y 2 and C x 3- y 3 are the vertices of a triangle- then the excentre opposite to B isA- None of theseB- a x1-b x2-c x3-a-b-c- a y1-b y2-c y3-a-b-cC- a x1-b x2-c x3-a-b-c- a y1-b y2-c y3-a-b-cD- a x1-b x2-c x3-a-b-c- a y1-b y2-c y3-a-b-c

The correct option is B It is obvious. ...

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Circumcentre

If A x 1, y 1...

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If $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ are the vertices of a triangle, then the excentre opposite to B is

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None of these

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If $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ are the vertices of a triangle, then the excentre with respect to B is

Find the centroid of the triangle with vertices A $(x_{1}, y_{1})$ , B $(x_{2}, y_{2})$ and C $(x_{3}, y_{3})$ .

A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})

∣ ∣ ∣ ∣ \begin{matrix} x & y & 1 x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} x & y & 1 x_{1} & y_{1} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0

A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})

A, B, C

\frac{1}{2} | Δ |

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

a = B C, b = C A, c = A B

2 s = a + b + c

Δ^{2}

A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})

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