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A variable line passing through a fixed point $(\alpha, \beta)$ intersects the coordinate axes at $A$ and $B$. If $O$ is the origin, then the locus of the centroid of the $\triangle O A B$ is
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Verified Answer
The correct answer is:
$\beta x+\alpha y-3 x y=0$
Let points $A(a, 0)$ and $B(0, b)$, so equation of variable line is
$$
\frac{x}{a}+\frac{y}{b}=1
$$
Since the variable line (i) passes through the point $(\alpha, \beta)$
So,
$$
\frac{\alpha}{a}+\frac{\beta}{b}=1
$$
Now, centroid of $\triangle O A B$ is $\left(\frac{a}{3}, \frac{b}{3}\right)=(h, k)$
So,
$$
a=3 h \text { and } b=3 k
$$
From Eqs. (ii) and (iii), we are getting
$$
\frac{\alpha}{3 h}+\frac{\beta}{3 k}=1
$$
On taking locus of point $(h, k)$, we are getting
$$
\beta x+\alpha y-3 x y=0 .
$$
$$
\frac{x}{a}+\frac{y}{b}=1
$$
Since the variable line (i) passes through the point $(\alpha, \beta)$
So,
$$
\frac{\alpha}{a}+\frac{\beta}{b}=1
$$
Now, centroid of $\triangle O A B$ is $\left(\frac{a}{3}, \frac{b}{3}\right)=(h, k)$
So,
$$
a=3 h \text { and } b=3 k
$$
From Eqs. (ii) and (iii), we are getting
$$
\frac{\alpha}{3 h}+\frac{\beta}{3 k}=1
$$
On taking locus of point $(h, k)$, we are getting
$$
\beta x+\alpha y-3 x y=0 .
$$
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