If l, m, n are in arithmetic progression,then | vedclass

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(B) Given that $l, m, n$ are in arithmetic progression $(AP)$. Therefore,$2m = l + n$. The equation of the line is $lx + my + n = 0$. We can rewrite t

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$(1, -2)$

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(B) Given that $l, m, n$ are in arithmetic progression $(AP)$. Therefore,$2m = l + n$. The equation of the line is $lx + my + n = 0$. We can rewrite the equation as $lx + my + n = 0$. Substituting $n = 2m - l$ into the equation: $lx + my + (2m - l) = 0$ $l(x - 1) + m(y + 2) = 0$ For this equation to be independent of $l$ and $m$,the coefficients must be zero: $x - 1 = 0 \implies x = 1$ $y + 2 = 0 \implies y = -2$ Thus,the line always passes through the point $(1, -2)$.

If $l, m, n$ are in arithmetic progression,then the straight line $lx + my + n = 0$ will always pass through the point:

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