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The Mid-Point Theorem

In any triang...

Question

In any triangle ABC, if equation of sides AB & AC are given by $3 x + 4 y - 7 = 0$ & $4 x - 3 y - 1 = 0$ & the mid point of BC is $(0, 0)$ , then centriod of $Δ A B C$ is

(\frac{2}{3}, - \frac{1}{3})

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(\frac{2}{3}, - \frac{2}{3})

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(1, 1)

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(\frac{1}{3}, \frac{1}{3})

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Solution

The correct option is D $(\frac{1}{3}, \frac{1}{3})$
$A B$ and $A C$ are given by $3 x + 4 y - 7 = 0$ & $4 x - 3 y - 1 = 0$
The intersection point would be $A$ .
Solving both the equation gives $A = (x, y) = (1, 1)$
Centroid divides the median in the ration $2 : 1$
$∴$ $c e n t r o i d = \frac{2 \times 0 + 1 \times 1}{2 + 1}, \frac{2 \times 0 + 1 \times 1}{2 + 1}$
$= (\frac{1}{3}, \frac{1}{3})$

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In any triangle ABC, if equation of sides AB & AC are given by 3x+4y−7=0 & 4x−3y−1=0 & the mid point of BC is (0,0), then centriod of ΔABC is

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