From the point (3,-4),perpendicular lines L_1 | vedclass

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(B) Given the pair of lines $S \equiv 2x^2+3xy-2y^2-7x+y+3=0$.Factorizing the homogeneous part: $2x^2+3xy-2y^2 = (x+2y)(2x-y) = 0$.Let the lines be $(

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$\frac{72}{5}$

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(B) Given the pair of lines $S \equiv 2x^2+3xy-2y^2-7x+y+3=0$.Factorizing the homogeneous part: $2x^2+3xy-2y^2 = (x+2y)(2x-y) = 0$.Let the lines be $(x+2y+c_1)(2x-y+c_2) = 2x^2+3xy-2y^2-7x+y+3$.Comparing coefficients of $x$ and $y$: $2c_1+c_2 = -7$ and $2c_2-c_1 = 1$.Solving these,we get $c_1 = -3$ and $c_2 = -1$.So,the lines are $x+2y-3=0$ and $2x-y-1=0$.The lines $L_1$ and $L_2$ are perpendicular to these lines and pass through $(3,-4)$.Line perpendicular to $x+2y-3=0$ is $2x-y+k_1=0$. Passing through $(3,-4)$: $2(3)-(-4)+k_1=0 \Rightarrow k_1=-10$. So,$2x-y-10=0$.Line perpendicular to $2x-y-1=0$ is $x+2y+k_2=0$. Passing through $(3,-4)$: $3+2(-4)+k_2=0 \Rightarrow k_2=5$. So,$x+2y+5=0$.The area of the rectangle formed by these four lines is the product of the distances between the parallel pairs.Distance between $x+2y-3=0$ and $x+2y+5=0$ is $d_1 = \frac{|5-(-3)|}{\sqrt{1^2+2^2}} = \frac{8}{\sqrt{5}}$.Distance between $2x-y-1=0$ and $2x-y-10=0$ is $d_2 = \frac{|-10-(-1)|}{\sqrt{2^2+(-1)^2}} = \frac{9}{\sqrt{5}}$.Area $= d_1 	imes d_2 = \frac{8}{\sqrt{5}} 	imes \frac{9}{\sqrt{5}} = \frac{72}{5} 	ext{ sq. units}$.

From the point $(3,-4)$,perpendicular lines $L_1$ and $L_2$ are drawn to each of the lines represented by $S \equiv 2x^2+3xy-2y^2-7x+y+3=0$. The area of the quadrilateral formed by the pair of lines $S=0$,$L_1$,and $L_2$ is (in square units):

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