Multiple choice

If $a,b$ and $c$ are position vector of $A,B$ and $C$ respectively of $\triangle ABC$ and if $|a-b|=4,|b-c|=2, |c-a|=3$, then the distance between the centroid and incentre of $\triangle ABC$ is

  1. $1$
  2. $\dfrac{1}{2}$
  3. $\dfrac{1}{3}$
  4. $\dfrac{2}{3}$
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A Correct answer
AI explanation

The side lengths of the triangle are given as |a-b| = 4, |b-c| = 2, and |c-a| = 3. By placing B at the origin (0,0) and C at (2,0), the coordinates for A are found using the distance formulas to be (9/4, 3 times the square root of 15 divided by 4). The centroid G is calculated as the average of the vertices' coordinates, resulting in (17/12, the square root of 15 divided by 4). The coordinates of the incentre I are found using the weighted average formula based on opposite side lengths, resulting in (5/3, 6 times the square root of 15 divided by 9). Calculating the distance between G and I using the distance formula yields the square root of (1/9), which equals 1/3. The correct answer is 1/3.

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