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…
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
\(6 x^2-9 x y+y^2-25 x+30 y+25=0\)
\(8 x^2-14 x y+3 y^2-25 x+30 y+50=0\)
\(8 x^2-14 x y+3 y^2-50 x+50 y+75=0\)
\(6 x^2+14 x y-3 y^2-30 x+40 y-75=0\)
Solution
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
\(\begin{array}{r}
(4 x-y-15)(2 x-3 y-5)=0 \\
\Rightarrow 8 x^2-14 x y+3 y^2-50 x+50 y+75=0
\end{array}\)