Let 3x^2+8xy-3y^2=0 represent the lines L_1, | vedclass

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(D) The given pair of lines are $S = 3x^2+8xy-3y^2 = 0$ and $S' = 3x^2+8xy-3y^2+2x-4y-1 = 0$.Let the intersection points be $P = L_1 \cap L_3$ and $Q

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$\frac{1}{16}$

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(D) The given pair of lines are $S = 3x^2+8xy-3y^2 = 0$ and $S' = 3x^2+8xy-3y^2+2x-4y-1 = 0$.Let the intersection points be $P = L_1 \cap L_3$ and $Q = L_2 \cap L_4$.The line $L$ passing through the intersection points of the two pairs of lines is given by $S' - S = 0$.Thus,$(3x^2+8xy-3y^2+2x-4y-1) - (3x^2+8xy-3y^2) = 0$.This simplifies to $2x - 4y - 1 = 0$.The intercepts of this line $L$ on the coordinate axes are found by setting $y=0$ and $x=0$ respectively.For $y=0$,$2x = 1 \Rightarrow x = \frac{1}{2}$. Point is $(\frac{1}{2}, 0)$.For $x=0$,$-4y = 1 \Rightarrow y = -\frac{1}{4}$. Point is $(0, -\frac{1}{4})$.The area of the triangle formed by the line with the coordinate axes is $\frac{1}{2} 	imes |	ext{base}| 	imes |	ext{height}|$.Area $= \frac{1}{2} 	imes |\frac{1}{2}| 	imes |-\frac{1}{4}| = \frac{1}{2} 	imes \frac{1}{2} 	imes \frac{1}{4} = \frac{1}{16}$ sq units.

Let $3x^2+8xy-3y^2=0$ represent the lines $L_1, L_2$ and $3x^2+8xy-3y^2+2x-4y-1=0$ represent the lines $L_3, L_4$. Let $L$ be the line joining the points of intersection of $L_1, L_3$ and $L_2, L_4$. Then,the area (in sq units) of the triangle formed by $L$ with the coordinate axes is

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