Every point (x, y) on the curve 3x + 2y - 3xy | vedclass

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(B) Let the line $(L)$ intersect the coordinate axes at $(a, 0)$ and $(0, b)$.The centroid of the triangle formed by the axes and the line $(L)$ is gi

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are concurrent

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(B) Let the line $(L)$ intersect the coordinate axes at $(a, 0)$ and $(0, b)$.The centroid of the triangle formed by the axes and the line $(L)$ is given by $(\frac{a}{3}, \frac{b}{3})$.Given that this point $(x, y)$ lies on the curve $3x + 2y - 3xy = 0$,we have $x = \frac{a}{3}$ and $y = \frac{b}{3}$,which implies $a = 3x$ and $b = 3y$.The equation of the line $(L)$ in intercept form is $\frac{X}{a} + \frac{Y}{b} = 1$.Substituting $a = 3x$ and $b = 3y$,we get $\frac{X}{3x} + \frac{Y}{3y} = 1$,or $\frac{X}{x} + \frac{Y}{y} = 3$.Since $(x, y)$ satisfies $3x + 2y - 3xy = 0$,we can write $3xy = 3x + 2y$,or $\frac{1}{y} + \frac{2}{3x} = 1$.Rewriting the line equation: $X(\frac{1}{x}) + Y(\frac{1}{y}) = 3$.Substituting $\frac{1}{y} = 1 - \frac{2}{3x}$,we get $X(\frac{1}{x}) + Y(1 - \frac{2}{3x}) = 3$.$X(\frac{1}{x}) + Y - Y(\frac{2}{3x}) = 3$.Multiplying by $3x$: $3X + 3xY - 2Y = 9x$.$3x(Y - 3) = 2Y - 3X$.$x = \frac{2Y - 3X}{3(Y - 3)}$.As $x$ and $y$ vary,the lines $\frac{X}{3x} + \frac{Y}{3y} = 1$ all pass through a fixed point. By testing values,we find they are concurrent at the point $(2, 3)$.

Every point $(x, y)$ on the curve $3x + 2y - 3xy = 0$ is the centroid of a triangle formed by the coordinate axes and a line $(L)$ intersecting both the coordinate axes. Then all such lines $(L)$

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