If I is the center of a circle inscribed in a triangle ABC- then -BC-IA-CA-IB-AB-IC is

We know that the position vector of the incenter of ΔABC is given by BC·a+CA·b+AB·cBC+CA+AB Where a, b, c are the affixes af the vertices of the triangle. There ...

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Standard X

Mathematics

Section Formula

If I is the c...

Question

If $I$ is the center of a circle inscribed in a triangle $A B C,$ then $∣ ∣ ∣ - - \to B C ∣ ∣ ∣ - \to I A + ∣ ∣ ∣ - - \to C A ∣ ∣ ∣ - \to I B + ∣ ∣ ∣ - - \to A B ∣ ∣ ∣ - \to I C$ is

\to 0

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- \to I A + - \to I B + - \to I C

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\frac{- \to I A + - \to I B + - \to I C}{3}

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\frac{- \to I A + - \to I B + - \to I C}{2}

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Solution

The correct option is A $\to 0$
We know that the position vector of the incenter of $Δ A B C$ is given by
$\frac{B C \cdot \to a + C A \cdot \to b + A B \cdot \to c}{B C + C A + A B}$
Where $\to a, \to b, \to c$ are the affixes af the vertices of the triangle.
Therefore, in the given triangle $A B C,$ the affix of $I$ is given by $\frac{∣ ∣ ∣ - - \to B C ∣ ∣ ∣ - \to I A + ∣ ∣ ∣ - - \to C A ∣ ∣ ∣ - \to I B + ∣ ∣ ∣ - - \to A B ∣ ∣ ∣ - \to I C}{∣ ∣ ∣ - - \to B C ∣ ∣ ∣ + ∣ ∣ ∣ - - \to C A ∣ ∣ ∣ + ∣ ∣ ∣ - - \to A B ∣ ∣ ∣}$
But it is the position vector of $I$ with respect to $\to O .$
$∴ \to O = \frac{∣ ∣ ∣ - - \to B C ∣ ∣ ∣ - \to I A + ∣ ∣ ∣ - - \to C A ∣ ∣ ∣ - \to I B + ∣ ∣ ∣ - - \to A B ∣ ∣ ∣ - \to I C}{∣ ∣ ∣ - - \to B C ∣ ∣ ∣ + ∣ ∣ ∣ - - \to C A ∣ ∣ ∣ + ∣ ∣ ∣ - - \to A B ∣ ∣ ∣}$
$\Rightarrow ∣ ∣ ∣ - - \to B C ∣ ∣ ∣ - \to I A + ∣ ∣ ∣ - - \to C A ∣ ∣ ∣ - \to I B + ∣ ∣ ∣ - - \to A B ∣ ∣ ∣ - \to I C = \to 0$

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If I is the center of a circle inscribed in a triangle ABC, then ∣∣∣−−→BC∣∣∣−→IA+∣∣∣−−→CA∣∣∣−→IB+∣∣∣−−→AB∣∣∣−→IC is

If $I$ is the center of a circle inscribed in a triangle $A B C,$ then $∣ ∣ ∣ - - \to B C ∣ ∣ ∣ - \to I A + ∣ ∣ ∣ - - \to C A ∣ ∣ ∣ - \to I B + ∣ ∣ ∣ - - \to A B ∣ ∣ ∣ - \to I C$ is