The joint equation of the lines pair of lines passing through the point $(3,-2)$ and perpendicular to the…
The joint equation of the lines pair of lines passing through the point $(3,-2)$ and perpendicular to the lines $5 x^2+2 x y-3 y^2=0$ is
$3 x^2+2 x y+5 y^2+14 x+26 y+5=0$
$3 x^2+2 x y-5 y^2-14 x-26 y-5=0$
$3 x^2-2 x y-5 y^2-14 x-26 y+5=0$
$3 x^2-2 x y+5 y^2+14 x+26 y-5=0$
Solution
Joint equation of the liens passing through the point $\left(x_1, y_1\right)$ and perpendicular to the lines $\mathrm{ax}^2+2 \mathrm{~h} x y+\mathrm{b} y^2=0$ is:
$\begin{aligned}
& \mathrm{a} x^2+2 \mathrm{~h} x y+\mathrm{b} y^2=0 \text { is: } \\
& \mathrm{b}\left(x-x_1\right)^2-2 \mathrm{~h}\left(x-x_1\right)\left(y-y_1\right)+\mathrm{a}\left(y-y_1\right)^2=0
\end{aligned}$
$\therefore \quad$ Equation of the required line is:
$\begin{array}{ll}
\therefore \quad & \text { Equation of the required line is: } \\
& -3(x-3)^2-2(x-3)(y+2)+5(y+2)^2=0 \\
\therefore \quad & -3\left(x^2-6 x+9\right)-2(x y+2 x-3 y-6) \\
& +5\left(y^2+4 y+4\right)=0
\end{array}$
$\begin{aligned} & \therefore \quad-3 x^2+18 x-27-2 x y-4 x+6 y+12+5 y^2 \\ & \quad+20 y+20=0 \\ & \therefore \quad 3 x^2+2 x y-5 y^2-14 x-26 y-5=0\end{aligned}$