The joint equation of pair of lines passing through point of intersection of lines
\(2 x^{2}-x y-15 y^{2}-7 x+32 y-9=0\) and parallel to co-ordinate axes is

Let \(\phi=2 x^{2}-x y-15 y^{2}-7 x+32 y-9=0\) ...(1)
\(\quad \frac{d \phi}{d x}=4 x-y-7=0 \Rightarrow 4 x-y-7=0\) ...(2)
\(\quad \frac{d \phi}{d y}=-x-30 y+32=0 \Rightarrow x+30 y-32=0\) ...(3)
Solving equation \((2) \&(3)\) we get \(x=2, y=1\)
\(\therefore(2,1)\) is the point of intersection of given lines.
Lines passing through \((2,1)\) and parallel to co-ordinate axes are \(x=2\) and \(y=1\).
Hence required equation is
\(\quad(x-2)(y-1)=0 \Rightarrow x y-2 y-x+2=0\)
Note : Point of intersection can also be calculated as follows :
\(2 x^{2}-x y-15 y^{2}-7 x+32 y-9=0 \text { gives } a=2, h=\frac{-1}{2},\) \( b=-15, g=\frac{-7}{2}, f=16, c=-9\)
\(\text { Point of intersection } =\left(\frac{\mathrm{bg}-\mathrm{hf}}{\mathrm{h}^{2}-\mathrm{ab}}, \frac{\mathrm{af}-\mathrm{gh}}{\mathrm{h}^{2}-\mathrm{ab}}\right) \)
\( \equiv\left[\frac{(-15)\left(-\frac{7}{2}\right)-\left(-\frac{1}{2}\right)(16)}{\left(-\frac{1}{2}\right)^{2}-(2)(-15)}, \frac{(2)(16)-\left(-\frac{7}{2}\right)\left(-\frac{1}{2}\right)}{\left(-\frac{1}{2}\right)^{2}-(2)(-15)}\right], \)
\( \equiv\left[\frac{\frac{105}{2}+\frac{16}{2}}{\frac{1}{4}+30}, \frac{32-\frac{7}{4}}{\frac{1}{4}+30}\right] \equiv\left[\frac{121}{2} \times \frac{4}{121}, \frac{121}{4} \times \frac{4}{121}\right] \)
\( \equiv(2,1) \)