A pair of lines $S=\mathbf{0}$ 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

Question

A pair of lines $S=\mathbf{0}$ 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, $6 x^2-9 x y+y^2-25 x+30 y+25=0$, $8 x^2-14 x y+3 y^2-25 x+30 y+50=0$, $8 x^2-14 x y+3 y^2-50 x+50 y+75=0$, $6 x^2+14 x y-3 y^2-30 x+40 y-75=0$

MARKS App · Accepted Answer

Equation of given pair of straight lines is $\begin{array}{rlrl} & 8 x^2-14 x y+3 y^2+10 x+10 y-25 & =0 \ \Rightarrow & & (4 x-y-5)(2 x-3 y+5) & =0 \end{array}$ Now point of intersection of lines $\begin{aligned} 4 x-y-5 & =0 \ 2 x-3 y+5 & =0 \text { is }(2,3) \end{aligned}$ So equation of $S=0$ is $\left(4 x-y+c_1\right)\left(2 x-3 y+c_2\right)=0$ and $S=0$ passes through a point $P\left(x_1, y_1\right)$ such that mid-point of $P\left(x_1, y_1\right)$ and $(2,3)$ is $(3,2)$. so, $\begin{array}{ll} \text {so, } & x_1=4 \text { and } y_1=1 \ \therefore & c_1=-15 \text { and } c_2=-5 \end{array}$ So, required equation is $\begin{array}{r} (4 x-y-15)(2 x-3 y-5)=0 \ \Rightarrow 8 x^2-14 x y+3 y^2-50 x+50 y+75=0 \end{array}$

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