If the lines represented by the equation 4xy | vedclass

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(C) The given equation is $4xy + 6x - 8y + c = 0$.We can rewrite this as $4x(y + \frac{6}{4}) - 8y + c = 0$,which simplifies to $4x(y + 1.5) - 8(y + 1

Vedclass · Accepted Answer

$\left|\frac{c}{4}\right|$

Vedclass · Answer

(C) The given equation is $4xy + 6x - 8y + c = 0$.We can rewrite this as $4x(y + \frac{6}{4}) - 8y + c = 0$,which simplifies to $4x(y + 1.5) - 8(y + 1.5) + 12 + c = 0$.This factors as $(4x - 8)(y + 1.5) + (12 + c) = 0$,or $(x - 2)(y + 1.5) = -\frac{12 + c}{4}$.For the lines to form a rectangle with the coordinate axes,the lines must be of the form $x = h$ and $y = k$.Comparing $(x - 2)(y + 1.5) = K$,where $K = -\frac{12 + c}{4}$,the lines are $x = 2$ and $y = -1.5$.The rectangle is formed by the lines $x = 0, x = 2, y = 0, y = -1.5$.The dimensions of the rectangle are $|2 - 0| = 2$ and $|-1.5 - 0| = 1.5$.The area is $2 	imes 1.5 = 3$.However,looking at the options,we check the condition for the lines to be $x = h$ and $y = k$ from $4xy + 6x - 8y + c = 0$.This implies $4(x - 2)(y + 1.5) = 12 - c$,so $(x - 2)(y + 1.5) = \frac{12 - c}{4}$.For the lines to be $x = 0$ and $y = 0$ as part of the rectangle,the constant term must be $0$,so $12 - c = 0$,meaning $c = 12$.The area is $|2 	imes (-1.5)| = 3$. Since $|c|/4 = 12/4 = 3$,the correct option is $C$.

If the lines represented by the equation $4xy + 6x - 8y + c = 0$ form a rectangle with the coordinate axes,then the area of the rectangle (in sq. units) is

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