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 factor this equation as follows:$4xy - 8y + 6x + c = 0$$4y(x - 2) + 6x + c = 0$$4y(x - 2) + 6

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$|c/4|$

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(C) The given equation is $4xy + 6x - 8y + c = 0$.We can factor this equation as follows:$4xy - 8y + 6x + c = 0$$4y(x - 2) + 6x + c = 0$$4y(x - 2) + 6(x - 2) + 12 + c = 0$$(4y + 6)(x - 2) = -(12 + c)$$(2y + 3)(x - 2) = -\frac{12 + c}{2}$For the lines to form a rectangle with the coordinate axes,the lines must be of the form $x = h$ and $y = k$.This implies the equation must be of the form $(x - h)(y - k) = 0$,which expands to $xy - kx - hy + hk = 0$.Comparing $4xy + 6x - 8y + c = 0$ with $4(xy + 1.5x - 2y + c/4) = 0$,we identify the lines as $x = 2$ 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 $2 	imes 1.5 = 3$.From the equation $(x - 2)(y + 1.5) = 0$,we have $xy + 1.5x - 2y - 3 = 0$,so $4xy + 6x - 8y - 12 = 0$.Comparing this with $4xy + 6x - 8y + c = 0$,we get $c = -12$.The area is $|c/4| = |-12/4| = 3$.

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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