A pair of lines S=0 together with the lines g | vedclass

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(C) The given pair of lines is $8 x^2-14 x y+3 y^2+10 x+10 y-25=0$.Factoring the expression,we get $(4 x-y-5)(2 x-3 y+5)=0$.The lines are $L_1: 4 x-y-

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$8 x^2-14 x y+3 y^2-50 x+50 y+75=0$

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(C) The given pair of lines is $8 x^2-14 x y+3 y^2+10 x+10 y-25=0$.Factoring the expression,we get $(4 x-y-5)(2 x-3 y+5)=0$.The lines are $L_1: 4 x-y-5=0$ and $L_2: 2 x-3 y+5=0$.The intersection point of $L_1$ and $L_2$ is $(2,3)$.Let the pair of lines $S=0$ be $(4 x-y+c_1)(2 x-3 y+c_2)=0$.Since these lines form a parallelogram with $L_1$ and $L_2$,the diagonals intersect at $(3,2)$.The intersection of the new pair of lines is $(x_1, y_1)$. The midpoint of $(2,3)$ and $(x_1, y_1)$ is $(3,2)$.Thus,$\frac{x_1+2}{2}=3 \Rightarrow x_1=4$ and $\frac{y_1+3}{2}=2 \Rightarrow y_1=1$.Substituting $(4,1)$ into the lines $4 x-y+c_1=0$ and $2 x-3 y+c_2=0$,we get $c_1=-15$ and $c_2=-5$.The equation $S=0$ is $(4 x-y-15)(2 x-3 y-5)=0$.Expanding this,we get $8 x^2-14 x y+3 y^2-50 x+50 y+75=0$.

$A$ pair of lines $S=0$ together with the lines given by the equation $8 x^2-14 x y+3 y^2+10 x+10 y-25=0$ form a parallelogram. If its diagonals intersect at the point $(3,2)$,then the equation $S=0$ is

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