A variable line passing through a fixed point | vedclass

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(B) Let the points be $A(a, 0)$ and $B(0, b)$. The equation of the variable line is $\frac{x}{a} + \frac{y}{b} = 1$. Since the line passes through $(\

Vedclass · Accepted Answer

$\beta x + \alpha y - 3 xy = 0$

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(B) Let the points be $A(a, 0)$ and $B(0, b)$. The equation of the variable line is $\frac{x}{a} + \frac{y}{b} = 1$. Since the line passes through $(\alpha, \beta)$,we have $\frac{\alpha}{a} + \frac{\beta}{b} = 1$. The centroid $(h, k)$ of $	riangle OAB$ is given by $h = \frac{a+0+0}{3} = \frac{a}{3}$ and $k = \frac{0+b+0}{3} = \frac{b}{3}$. Thus,$a = 3h$ and $b = 3k$. Substituting these into the line equation: $\frac{\alpha}{3h} + \frac{\beta}{3k} = 1$. Multiplying by $3hk$,we get $\alpha k + \beta h = 3hk$. Replacing $(h, k)$ with $(x, y)$,the locus is $\beta x + \alpha y = 3xy$,or $\beta x + \alpha y - 3xy = 0$.

$A$ 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 $\triangle OAB$ is

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